Skip to main content

General Physics I:

Sunil Karna

Section 10.7 Examples B

KTG.

Example 10.7.1.

The rms speed of \(H_{2} \) molecules is 1.84 km/sec. What will be the rms speed of \(O_{2} \) molecules at the same temperature? The molecular weights of hydrogen and oxygen are 2 and 32 respectively.
Solution.
We know that
\begin{equation*} v_{rms} = \sqrt{\frac{3RT}{M}} \end{equation*}
\begin{equation*} \text{or,}\quad \frac{(v_{rms})_{O}}{(v_{rms})_{H}} = \sqrt{\frac{M_{H}}{M_{O}}} \end{equation*}
\begin{equation*} \therefore\quad (v_{rms})_{O} = (v_{rms})_{H}\times \sqrt{\frac{M_{H}}{M_{O}}} =(1.84 km/s)\times \sqrt{\frac{2}{32}} =0.459 \,km/s. \end{equation*}

Example 10.7.2.

At what temperature does the kinetic energy of an atom be 1.0 eV? \((k_{B}=1.38\times 10^{-23} \,J/K)\text{.}\)
Solution.
Average kinetic energy \(e_{k}\) of an atom is given by
\begin{equation*} \frac{1}{2}mv_{rms}^{2}=\frac{f}{2}k_{B}T = \frac{3}{2}k_{B}T \end{equation*}
\(\because f=3\) for a monoatomic molecule
\begin{equation*} \therefore\quad T = \frac{2}{3}\frac{e_{k}}{k_{B}} \end{equation*}
\begin{equation*} = \frac{2}{3}\left(\frac{1.0\, eV\times 1.60\times 10^{-19} \,J/eV}{1.38\times 10^{-23} \,J/K}\right)= 7.729\times 10^{3} \,K \end{equation*}

Example 10.7.3.

At what temperature will the rms velocity of hydrogen molecules has double of its value at NTP, if pressure remains constant.
Solution.
We have -
\begin{equation*} v_{rms}=\sqrt{\frac{3k_{B}T}{m}} \end{equation*}
At NTP, \(T=0^{o}C=273 \,K\text{.}\)
\begin{equation*} (v_{rms})_{1}=v_{rms}=\sqrt{\frac{3k_{B}\times 273}{m}} \end{equation*}
At NTP, \(T=T \,K\)
\begin{equation*} (v_{rms})_{2}=2v_{rms} = \sqrt{\frac{3k_{B}\times T}{m}} \end{equation*}
\begin{equation*} \text{or,}\quad 2v_{rms} =\sqrt{\frac{3k_{B}\times T}{m}} = 2\sqrt{\frac{3k_{B}\times 273}{m}} \end{equation*}
\begin{equation*} \text{or,}\quad \frac{T}{273} = 4 \end{equation*}
\begin{equation*} \therefore\quad T = 4\times 273=1092 \,K = 819 \,^{o}C \end{equation*}

I Law of Thermodynamics.

Example 10.7.4.

A spherical balloon one foot in diameter contains helium at one atmosphere of pressure and at a temperature of \(27\,^{o}C\text{.}\) Find
  1. the number of moles of helium in the balloon,
  2. the number of molecules in the balloon,
  3. the average kinetic energy of each helium molecule,
  4. the rms velocity of a helium molecule,
  5. the amount of heat (in Joules) to increase the temperature of the helium in the balloon by one degree Celsius if the balloon is made of rubber that allows the pressure to remain at one atmosphere,
  6. the amount of heat (in Joules) to increase the temperature of the helium in the balloon by one degree Celsius if the balloon is made of non-stretching aluminum so the volume remains constant,
  7. the final temperature of the balloon if it rises adiabatically to an altitude where the pressure is one-half atmosphere.
Solution.
Given: \(r=0.5\,ft=0.1524 \,m,\quad p=1\,atm=1.013\times 10^{5} \,N/m^{2},\) \(T=27\,^{o}C=300 \,K, \quad R=8.314 \,J/mol.K\)
  1. \begin{equation*} pV=nRT \quad \Rightarrow\quad n=\frac{pV}{RT} \end{equation*}
    Here,
    \begin{equation*} V=\frac{4}{3}\pi r^{3} = \frac{4}{3}\pi (0.1524 \,m)^{3} =0.01483 \,m^{3} \end{equation*}
    \begin{equation*} \therefore\quad n=\frac{pV}{RT} =\frac{(1.013\times 10^{5} \,N/m^{2})(0.01483 \,m^{3})}{(8.314 \,J/mol.K)(300 \,K)} =0.602 \,mol. \end{equation*}
  2. \begin{equation*} N_{A} = 6.022\times 10^{23} \,molecules/mol, \end{equation*}
    \begin{equation*} N=nN_{A} =6.210\times 10^{23} \,molecules \end{equation*}
  3. \begin{equation*} k_{B} = 1.380\times 10^{-23}\,J/K, \end{equation*}
    \begin{equation*} e_{k} = \frac{3}{2}k_{B}T = 6.210\times 10^{-21}\,J \end{equation*}
  4. \(M=4.003 \,g/mol,\)
    \begin{equation*} e_{k} = \frac{1}{2}mv_{rms}^{2}, \end{equation*}
    \begin{equation*} m=\frac{M}{N_{A}} = 6.647\times 10^{-24} \,kg \end{equation*}
    \begin{equation*} \therefore\quad v_{rms} = \sqrt{\frac{2e_{k}}{m}} \end{equation*}
    \begin{equation*} = \sqrt{\frac{2(6.210\times 10^{-21}\,J)}{6.647\times 10^{-24} \,kg}}=43.22 \,m/s \end{equation*}
  5. \(\Delta T = 1\,^{o}C = 1 \,K,\)
    \begin{equation*} Q=nC_{p}\Delta T, \end{equation*}
    \begin{equation*} C_{p}=\frac{5}{2}R = 20.79 \,J/mol.K \end{equation*}
    \begin{equation*} \therefore\quad Q = (0.6022 \,mol)(20.79 \,J/mol.K)1\,K=12.52 \,J \end{equation*}
  6. \begin{equation*} Q=nC_{V}\Delta T, \end{equation*}
    \begin{equation*} C_{V}=\frac{3}{2}R =12.47 \,J/mol.K \end{equation*}
    \begin{equation*} \therefore\quad Q= (0.6022 \,mol)(12.47 \,J/mol.K) 1\,K=7.510 \,J \end{equation*}
  7. \begin{equation*} p_{2} =\frac{1}{2} \,atm = 5.065\times 10^{4} \,Pa, \end{equation*}
    \begin{equation*} pV^{\gamma} = p_{2}V_{2}^{\gamma}, \end{equation*}
    \begin{equation*} V_{2}=\left(\frac{pV^{\gamma}}{p_{2}}\right)^{\frac{1}{\gamma}}, \end{equation*}
    \begin{equation*} \gamma=\frac{5}{3} \end{equation*}
    \begin{equation*} \therefore\quad V_{2} = 0.02248 \,m^{3} \end{equation*}
    Now,
    \begin{equation*} p_{2}V_{2}=nRT_{2} \end{equation*}
    \begin{equation*} \Rightarrow\quad T_{2}=\frac{p_{2}V_{2}}{nR} =227.4\,K=-45.6\,^{o}C. \end{equation*}

Example 10.7.5.

Use the principle of equipartition of energy to determine the equations for the internal energy, the molar specific heat capacity at constant volume, and the molar specific heat capacity at constant pressure for
  1. a monatomic ideal gas,
  2. diatomic ideal gas,
  3. diatomic ideal gas with vibration, and
  4. solid substance.
Solution.
Internal energy per molecule of a gas,
\begin{equation*} U_{i}=\frac{f}{2}k_{B}T. \end{equation*}
\begin{equation*} \therefore \quad U=\frac{f}{2}Nk_{B}T =\frac{f}{2}nRT \end{equation*}
\begin{equation*} \text{or,}\quad \Delta U = \frac{f}{2}nR\Delta T=\Delta Q = nC_{V}\Delta T \end{equation*}
\begin{equation*} \therefore\quad C_{V} = \frac{f}{2}R \end{equation*}
  1. for a monoatomic gas, f = 3 translational motion = 3 degrees of freedom.
    \begin{equation*} \therefore\quad C_{V} = \frac{f}{2}R = \frac{3}{2}R \end{equation*}
  2. for a diatomic gas, f = 3 translational + 2rotational motion = 5 degrees of freedom.
    \begin{equation*} \therefore\quad C_{V} = \frac{f}{2}R = \frac{5}{2}R \end{equation*}
  3. for a diatomic gas with vibration, f = 3 translational + 2rotational + 2 vibrational motions = 7 degrees of freedom. (in 2 vibrational motions we have: 1 dimensional stretch and 1 dimensional bend)
    \begin{equation*} \therefore\quad C_{V} = \frac{f}{2}R = \frac{7}{2}R \end{equation*}
  4. for solids, f = 3 D kinetic energy + 3 D potential energy = 6 degrees of freedom.
    \begin{equation*} \therefore\quad C_{V} = \frac{f}{2}R = \frac{6}{2}R =3R \end{equation*}

Example 10.7.6.

An ideal monatomic gas containing 10 moles of Helium initially at \(0\,^{o}C\) and one atmosphere is allowed to expand to twice its initial volume. If the temperature remains constant throughout the process, find each of the following:
  1. the initial volume of the gas,
  2. the final volume of the gas,
  3. the final temperature of the gas,
  4. the final pressure of the gas,
  5. the work done by the gas,
  6. the change in internal energy of the gas, and
  7. the heat added to the gas.
Solution.
Given: \(n=10 \,mol,\quad T_{o}=0\,^{o}C=273 \,K,\) \(p_{o} =1.013\times 10^{5} \,Pa,\) \(R=8.314 \,J/mol-K,\quad V=2V_{o}, \) and \(T=T_{o}.\)
  1. \begin{equation*} p_{o}V_{o}=nRT_{o} \end{equation*}
    \begin{equation*} \therefore\quad V_{o} = \frac{nRT_{o}}{p_{o}} = 0.224 \,m^{3} \end{equation*}
  2. \begin{equation*} V=2V_{o} = 2\times 0.224 \,m^{3} = 0.448 \,m^{3} \end{equation*}
  3. \begin{equation*} T=T_{o} = 273 \,K \end{equation*}
  4. \begin{equation*} pV=nRT \end{equation*}
    \begin{equation*} \text{or,}\quad p(2V_{o})= nRT_{o} \end{equation*}
    \begin{equation*} \Rightarrow \quad p = \frac{nRT_{o}}{2V_{o}} = \frac{p_{o}}{2}=5.065\times 10^{4} \,Pa \end{equation*}
  5. \begin{equation*} W =\int p\,dV = \int\left(\frac{nRT_{o}}{V}\right)\,dV \end{equation*}
    \begin{equation*} = nRT_{o}\ln\left(\frac{V}{V_{o}}\right) = 1.573 \times 10^{4} \,J \end{equation*}
  6. \begin{equation*} \Delta U = \frac{3}{2}nR(T-T_{o}) =0 \end{equation*}
  7. \begin{equation*} \Delta U = Q- W \end{equation*}
    \begin{equation*} \therefore\quad Q = \Delta U +W = 1.573\times10^{4} \,J \end{equation*}

Example 10.7.7.

Calculate the change in the internal energy when 5 gm of air is heated from \(0 \,^{o}C\) to \(2\,^{o}C,\) the specific heat of air at constant volume being \(0.172 \,cal/gm/^{o}C.\) Note: 1 cal = 4.18 J.
Solution.
From first law of thermodynamics,
\begin{equation*} \,dQ=\,dU+p\,dV \end{equation*}
Since air is heated at constant volume \(\,dV=0.\)
\begin{equation*} \therefore \,dQ=\,dU \end{equation*}
\begin{equation*} \,dQ = mc_{v}\,dT \end{equation*}
\begin{equation*} =\,dU = (5 \,g) (0.172 \frac{cal}{g.^{o}C})(4.18 \frac{J}{cal}) (2-0)\,^{o}C = 7.19 \,J \end{equation*}

Example 10.7.8.

A system absorbs 2500 J of heat and at the same time does 90 J of external work.
  1. Find the change in internal energy of the system.
  2. Find the change in internal energy of the system if it absorbs 2500 J of heat at the same time 90 J of work done on it.
  3. What will be the change in internal energy of the gas from which 400 J of heat is removed at constant volume.
Solution.
From first law of thermodynamics,
\begin{equation*} \,dU=\,dQ-\,dW \end{equation*}
  1. Given: \(\,dQ= 2500 \,J,\quad \,dW=90\,J \)
    \begin{equation*} \therefore \,dU=2500-90 = 2410 \,J \end{equation*}
    We know that \(\,dW=p\,dV\text{,}\) hence if volume increases or \(\,dV\) is +ve, then \(\,dW\) is positive. That is, work done by the system is +ve.
  2. Given: \(\,dQ= 2500 \,J,\quad \,dW=-90 \,J \)
    \begin{equation*} \therefore \,dU=2500-(-90) = 2590 \,J \end{equation*}
    We know that \(\,dW=p\,dV\text{,}\) hence if volume decreases or \(\,dV \) is -ve, then \(\,dW\) is negative. That is, work done on the system is -ve.
  3. Given: \(\,dQ= -400 \,J \) and \(\,dV=0\)
    \begin{equation*} \therefore \,dU=\,dQ-p\,dV = -400 -0 = -400 \,J \end{equation*}
    -ve sign indicates that heat is taken out from the system.

II Law of Thermodynamics.

Example 10.7.9.

A heat engine takes in 1000 J of heat from a reservoir at temperature 400 K and rejects 800 J of heat into a second heat reservoir at temperature 300 K.
  1. How much work does the engine do per cycle.
  2. If the engine makes 4000 rpm, what is the power output of the engine in Watts? In horsepower?
  3. What is the actual efficiency of the engine?
  4. What would be the efficiency of a Carnot engine utilizing the same temperatures?
  5. Assuming the Carnot engine absorbed the same amount of heat per cycle as the original engine, how much work would it do per cycle?
  6. Assuming it ran at the same speed as the original, what would be its power output?
Solution.
Given: \(Q_{h} =1000 \,J, \quad T_{h} =400\,K,\) \(Q_{c}=800 \,J, \) and \(T_{c}= 300 \,K\)
  1. \begin{equation*} W= Q_{h} - Q_{c} = 200 \,J \end{equation*}
  2. \begin{equation*} p=\frac{W}{t} = Wf \end{equation*}
    \begin{equation*} \text{frequency},\quad f = 4000 \,rev/min = 66.67 \,rev/s \end{equation*}
    \begin{equation*} \therefore\quad p= 200 (J) \times 66.67 (rev/s) \end{equation*}
    \begin{equation*} = 13334 \,watt = 13.3 \,kW = 17.9 \,hp \end{equation*}
    \([\because 1hp=746 watt]\)
  3. \begin{equation*} \eta = \frac{W}{Q_{h}}\times 100\% = 20\% \end{equation*}
  4. \begin{equation*} \eta_{C} = \left(1-\frac{T_{c}}{T_{h}}\right)\times 100\% = 25\% \end{equation*}
  5. \begin{equation*} \eta_{C} = \frac{W_{C}}{Q_{h}} \end{equation*}
    \begin{equation*} \quad \Rightarrow\quad W_{C}= \eta_{C}\times Q_{h}= 0.25 \times 1000 (J) = 250 \,J \end{equation*}
  6. \begin{equation*} p_{C} = W_{C}f = 16.7 \,kW \end{equation*}

Example 10.7.10.

A Carnot engine is used to make a heat pump. In winter time, the engine pumps heat from outside the house at a temperature of \(0\,^{o}C\) into the house at a temperature \(30\,^{o}C\text{.}\)
  1. What is the COP of the heat pump?
  2. How much heat is delivered into the house in 24 hours if the unit uses 1000 W of electrical power? In the summer time, the engine is used as an air conditioner pumping heat from an exterior temperature of \(35\,^{o}C\) to an interior temperature of \(25\,^{o}C\text{.}\)
  3. What is the COP of this heat pump operating as an air conditioner?
  4. What is the cooling capacity of this unit in Btu’s per hour if it uses 1000 W of electrical power?
Solution.
  1. Given: \(T_{h}= 30\,^{o}C = 303 \,K, \quad T_{c}= 0\,^{o}C = 273 \,K.\)
    \begin{equation*} COP = \frac{T_{h}}{T_{h}-T_{c}} = 10.1 \end{equation*}
  2. Given: \(p=1000 \,W, \quad t=24 \,h= 24\times 60\times 60 \,s = 86400 \,s. \)
    \begin{equation*} COP = \frac{Q_{h}}{W} = \frac{Q_{h}}{pt} \end{equation*}
    \begin{equation*} \therefore\quad Q_{h} = COP (pt) \end{equation*}
    \begin{equation*} = 10.1\times (1000 \,Watt) \times (86400 \,s) =8.73\times 10^{8} \,J \end{equation*}
  3. Given: \(T_{h}= 35\,^{o}C = 308 \,K, \quad T_{c}= 25\,^{o}C = 298 \,K.\)
    \begin{equation*} COP = \frac{T_{h}}{T_{h}-T_{c}} = 29.8 \end{equation*}
  4. Given: \(p=1000 \,W, \quad t=1h= 3600 \,s.\)
    \begin{equation*} W=pt =1000 \,W\times 3600 \,s = 3.6\times 10^{6} \,J \end{equation*}
    \begin{equation*} COP = \frac{Q_{h}}{W} \end{equation*}
    \begin{equation*} \Rightarrow\quad Q_{c} = COP\times W = 29.8 \times 3.6\times 10^{6} \,J = 1.07\times 10^{8} \,J \end{equation*}
    \begin{equation*} \therefore\quad Q_{c} = \frac{1.07\times 10^{8} \,J}{1055 (J/Btu)} =102 \,k\,Btu \end{equation*}
    \begin{equation*} i.e.,\quad Q_{c}/t =102 \,k\,Btu/h \end{equation*}

Example 10.7.11.

An engine works between the temperature \(227\,^{o}C\) and \(27\,^{o}C \) and develops 100 horse power. Assuming that its efficiency is equal to that of a Carnot engine working between the same temperature limits, calculate:
  1. the heat supplied to the engine,
  2. the heat rejected by the engine.
Solution.
  1. The efficiency of Carnot heat engine is given by
    \begin{equation*} \eta = 1-\frac{T_{c}}{T_{h}} = 1-\frac{27+273}{227+273} = 1-\frac{300}{500} =0.4=40\% \end{equation*}
    Heat supplied by the engine per second is
    \begin{equation*} Q_{h}/t = \frac{W}{t\eta} = \frac{P}{\eta} \end{equation*}
    \begin{equation*} = \frac{(100\times 746 \,watt/sec)}{0.4} =186500 \,J/sec= 4.44\times 10^{4} \,cal/sec. \end{equation*}
  2. The efficiency of heat engine is given by
    \begin{equation*} \eta = 1-\frac{Q_{c}}{Q_{h}} =0.4 \end{equation*}
    \begin{equation*} \text{or,}\quad \frac{Q_{c}}{Q_{h}} =1-0.4 =0.6 \end{equation*}
    \begin{equation*} \therefore\quad Q_{c} = Q_{h}\times 0.6 \end{equation*}
    \begin{equation*} = (4.44\times 10^{4} \,cal/sec)\times 0.6 = 2.66\times 10 ^{4} \,cal/sec. \end{equation*}

Example 10.7.12.

A Carnot engine works at high temperature 600 K with the efficiency of \(40\%\text{.}\) If the efficiency of the engine is \(75\%\) and the low temperature kept constant, what is the high temperature?
Solution.
For Carnot engine efficiency,
\begin{equation*} \eta = 1-\frac{T_{c}}{T_{h}} \end{equation*}
\begin{equation*} \text{or,}\quad 0.4 = 1-\frac{T_{c}}{600} \end{equation*}
\begin{equation*} \therefore\quad T_{c} = 600(1-0.4)= 600\times 0.6 =360 \,K \end{equation*}
For heat engine efficiency,
\begin{equation*} \eta = 1-\frac{T_{c}}{T_{h}} \end{equation*}
\begin{equation*} \text{or,}\quad 0.75 = 1-\frac{360}{T_{h}} \end{equation*}
\begin{equation*} \text{or,}\quad \frac{360}{T_{h}} = 1-0.75=0.25 \end{equation*}
\begin{equation*} \therefore\quad T_{h} = \frac{360}{0.25} =1440 \,K \end{equation*}

Example 10.7.13.

  1. A 10 moles of \(O_{2}\) gas are allowed to expand isothermally at T = 300 K from an initial volume of 10 liters to a final volume of 30 liters. How much work does the gas do on the piston?
  2. If the gas expands adiabatically from 10 to 30 liters. How much work does the gas do?
Solution.
  1. Work done by the gas in an isothermal process
    \begin{equation*} W=\int\limits_{v_{i}}^{v_{f}} p\,dV =nRT\int\limits_{v_{i}}^{v_{f}} \frac{\,dV}{V}=nRT\ln\frac{V_{f}}{V_{i}} \end{equation*}
    \begin{equation*} \therefore W= 10\times 8.314 \times 300 \ln(3) = 2.7\times 10^{4} \,J \end{equation*}
    Since n = 10 mole, R = 8.314 J/mol/K, \(V_{i}=10 \,L,\) and \(V_{f}=30 \,L,\text{.}\)
  2. Work done by the gas in an adiabatic process
    \begin{equation*} W=\int\limits_{v_{i}}^{v_{f}} p\,dV =k \int\limits_{v_{i}}^{v_{f}} \frac{\,dV}{V^{\gamma}} \end{equation*}
    \begin{equation*} = \frac{k}{1-\gamma}\left[V^{1-\gamma}\right]_{v_{i}}^{v_{f}} \end{equation*}
    but,
    \begin{equation*} k=p_{i}V_{i}^{\gamma} =\frac{nRT}{V_{i}}V_{i}^{\gamma} \end{equation*}
    \begin{equation*} =nRTV_{i}^{\gamma-1} = 10\times 8.314\times 300\times (0.01)^{\frac{7}{5}-1} = 3953\quad \text{SI units} \end{equation*}
    where n = 10 mole, R = 8.314 J/mol/K, \(V_{i}=10 \,L = 0.01 \,m^{3},\) and \(\gamma=\frac{7}{5}\) for diatomic molecule.
    \begin{equation*} \text{or,}\quad W = \frac{3953}{1-\frac{7}{5}} \left[V_{f}^{1-\frac{7}{5}}-V_{i}^{1-\frac{7}{5}}\right] \end{equation*}
    \begin{equation*} = -\frac{3953\times 5}{2} \left[V_{f}^{-\frac{2}{5}}-V_{i}^{-\frac{2}{5}}\right] \end{equation*}
    \begin{equation*} \therefore \quad W= -9882\times \left[0.03^{-\frac{2}{5}}-0.01^{-\frac{2}{5}}\right] = 22230 \,J =2.2\times 10^{4} \,J \end{equation*}

Example 10.7.14.

If 3 moles of an ideal polyatomic gas initially with a volume of \(2 m^{3},\) and a temperature of 273 K. This gas is compressed isothermally to 1/2 its initial volume. How much heat must be added to the system during this compression?
Solution.
\begin{equation*} W=\int\limits_{v_{i}}^{v_{f}}p\,dV = \int\limits_{v_{i}}^{v_{f}} nRT\frac{\,dV}{V} = nRT\ln\frac{V_{f}}{V_{i}} \end{equation*}
\begin{equation*} = 3\times 8.314\times 273 \times \ln(\frac{1}{2})= -4719 \,J \end{equation*}
Now, at constant T, \(\,dU=0\text{.}\)
\begin{equation*} \therefore\quad \,dQ=\,dU+\,dW =0+ \,dW \end{equation*}
Hence,
\begin{equation*} Q = W = -4.7 \times 10^{4}\,J \end{equation*}

Example 10.7.15.

The molar heat capacity of hydrogen gas at constant volume is \(C_{v}=\frac{5}{2}R.\) It fills a volume of \(100 \,cm^{3}\) at pressure of 51 kPa. Determine the work done by the gas when it expands its volume five times
  1. isothermally, and
  2. adiabatically.
Solution.
  1. Work done in an isothermal process is given by
    \begin{equation*} W=\int\limits_{v_{i}}^{v_{f}}p\,dV =p_{i}V_{i}\int\limits_{v_{i}}^{v_{f}}\frac{\,dV}{V} = p_{i}V_{i} \ln\frac{V_{f}}{V_{i}} \end{equation*}
    \([\because pV = p_{i}V_{i}]\)
    \begin{equation*} \therefore\quad W = 51 \times 10^{3}(N/m^{2})(100\times 10^{-6}m^{3})\ln(5)=8.2 \,J \end{equation*}
  2. Work done in an adiabatic process is given by
    \begin{equation*} W=\int\limits_{v_{i}}^{v_{f}}p\,dV =p_{i}V_{i}^{\gamma}\int\limits_{v_{i}}^{v_{f}}\frac{\,dV}{V^{\gamma}} \end{equation*}
    \begin{equation*} = \frac{p_{i}V_{i}^{\gamma} }{1-\gamma}\left[V^{1-\gamma}\right]_{v_{i}}^{v_{f}} \end{equation*}
    \([\because pV^{\gamma}=p_{i}V_{i}^{\gamma}]\)
    \begin{equation} \text{or,}\quad W = \frac{p_{i}V_{i}}{1-\gamma}\left[5^{1-\gamma}-1\right] \tag{10.7.1} \end{equation}
    \([\because V_{f}=5V_{i}] \) Now,
    \begin{equation*} \gamma= \frac{C_{p}}{C_{v}} \end{equation*}
    \begin{equation*} \text{and}\quad C_{p}=C_{v}+R = \frac{7}{2}R \end{equation*}
    \begin{equation*} \therefore\quad \gamma = \frac{7}{5} \end{equation*}
    Hence from eqn. (10.7.1),
    \begin{equation*} W= \frac{p_{i}V_{i}}{1-\gamma}\left[5^{1-\gamma}-1\right] \end{equation*}
    \begin{equation*} = \frac{51\times 10^{3}\times 100\times 10^{-6}}{1-\frac{7}{5}}\left[5^{-2/5} -1\right] \end{equation*}
    \begin{equation*} \therefore W = 6.05\,J \end{equation*}

Entropy.

Example 10.7.16.

A cube of ice of mass 100 grams and temperature \(0\,^{o}C\) is placed in a beaker and heated until the ice melts and the water reaches \(100\,^{o}C\text{.}\) What is the change in entropy of the mass?
Solution.
Given: \(m=100 \,g,\quad T_{1}=0\,^{o}C = 273 \,K,\) \(T_{2}=100\,^{o}C = 373 \,K, \quad L=334\,J/g,\) and \(c=4.186 \,J/g-K.\) Change in entropy for ice melts -
\begin{equation*} \Delta S_{1} =\frac{\Delta Q_{1}}{T_{1}} \end{equation*}
\begin{equation*} = \frac{mL}{T_{1}} = \frac{100 \,g\times 334 \,J/g}{273\,K} = 122.3 \,J/K \end{equation*}
Change in entropy for water heating up -
\begin{equation*} \Delta S_{2} =\int \frac{\,dQ}{T} \end{equation*}
\begin{equation*} \text{but,}\quad \,dQ = mc\,dT \end{equation*}
\begin{equation*} \therefore\quad \Delta S_{2} =mc\int\limits_{T_{1}}^{T_{2}} \frac{\,dT}{T} \end{equation*}
\begin{equation*} = mc\ln\left(\frac{T_{2}}{T_{1}}\right) = 130.6 \,J/K \end{equation*}
Therefore total change in entropy,
\begin{equation*} \Delta S = \Delta S_{1}+\Delta S_{2} = 253 \,J/K. \end{equation*}

Example 10.7.17.

A piece of iron \((c = 448 \,J/kg.^{o}C)\) having a mass of 300 g and temperature of \(200\,^{o}C\) is placed in a well insulated and massless calorimeter containing 500 g of water \((c = 4186\,J/kg.^{o}C)\) at \(30\,^{o}C\text{.}\)
  1. What is the final temperature of the system?
  2. What is the change in entropy of the iron?
  3. What is the change in entropy of the water?
  4. What is the change in entropy of the universe?
  5. Is this result consistent with the second law of thermodynamics? Why?
Solution.
Given: \(m_{1}=300 \,g =0.3\,kg, \quad m_{2}=500 \,g =0.5\,kg, \) \(c_{1}=448 \,J/kg-K, \quad c_{2}=4186 \,J/kg-K,\) \(T_{1}=200\,^{o}C = 473 \,K \) and \(T_{2}=30\,^{o}C = 303 \,K.\)
  1. From the principle of calorimetry, Heat lost by iron = heat gained by water.
    \begin{equation*} m_{1}c_{1}(T_{1}-T_{f}) = m_{2}c_{2}(T_{f}-T_{2}) \end{equation*}
    \begin{equation*} \text{or,}\quad (0.3 \,kg)(448 \,J/kg.K) (473-T_{f}) = (0.5 \,kg)(4186 \,J/kg.K)(T_{f}-303) \end{equation*}
    \begin{equation*} \therefore\quad T_{f}= 313 \,K \end{equation*}
  2. \begin{equation*} \Delta S_{1} = \int\limits_{T_{1}}^{T_{2}} \frac{mc\,dT}{T} = mc\ln\left(\frac{T_{f}}{T_{1}}\right) = -55.4 \,J/K \end{equation*}
  3. \begin{equation*} \Delta S_{2} = \int\limits_{T_{2}}^{T_{f}} \frac{mc\,dT}{T} = mc\ln\left(\frac{T_{f}}{T_{2}}\right) = 69.7\,J/K \end{equation*}
  4. \begin{equation*} \Delta S = \Delta S_{1}+\Delta S_{2} =14.3 \,J/K \end{equation*}
  5. Yes, entropy of the universe increases.

Example 10.7.18.

On the microscopic level entropy is defined as
\begin{equation*} S = N k_{B} ln \omega\text{,} \end{equation*}
where \(\omega\) is the number of microscopic states available to the system in its macroscopic state. For a single gas molecule, the number of available microscopic states is proportional to the volume V of the container so that
\begin{equation*} \omega_{1} = k V \end{equation*}
where k is a constant of proportionality.
  1. If there are N such molecules in the same container, show that the number of available microscopic states is \(\omega = \omega_{1}^{N}\)
  2. How many microscopic states are available for one mole of gas in a volume V?
  3. In terms of \(k, \,V, \,N_{A},\) and \(k_{B}\text{,}\) what is the entropy of a mole of gas in volume V?
  4. If one mole of gas initially in a volume of 1 L is allowed to expand freely into a vacuum of volume 2 L, what is the change in entropy of the gas?
  5. Is this result consistent with the second law of thermodynamics? Why?
Solution.
  1. For each state \(\omega_{1} \) of one molecule a second molecule can be in any one of \(\omega_{2}\) states. Therefore, the total number of microstates for two particles is \(\omega_{2} = W_{1}^{2}.\) For each of these states the third particle cab be in any one of \(\omega_{1} \) states, so \(\omega_{3} = \omega_{1}^{3}. \) Generalizing to N particles gives, \(\omega_{N}=\omega_{1}^{N}.\)
  2. For one mole of gas, there are \(N_{A}\) molecules, so the above equation becomes,
    \begin{equation*} \omega = \omega_{1}^{N_{A}} = (kV)^{N_{A}} \end{equation*}
  3. Since \(S=Nk_{B}\ln \omega,\) then for one mole,
    \begin{equation*} S=N_{A}k_{B}\ln (kV)^{N_{A}} = N_{A}^{2}k_{B}\ln (kV) \end{equation*}
  4. \(V_{1}= 1L, \quad V_{2} =2L\)
    \begin{equation*} \Delta S = S_{2}-S_{1} = N_{A}^{2}k_{B}\ln (kV_{2}) - N_{A}^{2}k_{B}\ln (kV_{1}) \end{equation*}
    \begin{equation*} = N_{A}^{2}k_{B}\ln \left(\frac{V_{2}}{V_{1}}\right)= N_{A}^{2}k_{B}\ln 2 \end{equation*}
  5. Yes, entropy of the universe increases.

Example 10.7.19.

Calculate the change in entropy of 1 g of ice at \(0\,^{o}C\) is concerted into steam at \(100\,^{o}C\text{.}\) Latent heat of fusion = 80 cal/g, latent heat of vaporization = 540 cal/g, and specific heat capacity of water is \(1 \,cal/g/^{o}C.\)
Solution.
  1. When ice melts to water at \(0\,^{o}C,\) then change in entropy,
    \begin{equation*} \,dS_{1} = \frac{\,dQ}{T} = \frac{mL_{f}}{T} \end{equation*}
    \begin{equation*} = \frac{1\times 80}{273} =0.293 \,cal/K \end{equation*}
  2. When water is heated from \(0\,^{o}C \) to \(100\,^{o}C,\) then change in entropy,
    \begin{equation*} \,dS_{2} =\int \frac{\,dQ}{T} \end{equation*}
    \begin{equation*} = mc\ln \frac{T_{f}}{T_{i}} = 1\times 1\times \ln \frac{373}{273} =0.312 \,cal/K \end{equation*}
  3. When water is converted into steam at \(100\,^{o}C,\) then change in entropy,
    \begin{equation*} \,dS_{3} = \frac{\,dQ}{T} \end{equation*}
    \begin{equation*} = \frac{mL_{v}}{T} = \frac{1\times 540}{373} =1.448 \,cal/K \end{equation*}
    Hence the total change in entropy when one gram of ice at \(0\,^{o}C\) is converted into steam at \(100\,^{o}C\) is
    \begin{equation*} S=0.293+0.312+1.448 = 2.053 \,cal/K \end{equation*}

Example 10.7.20.

Calculate the change in entropy when 1 kg of water at \(0^{o}C\) is mixed with 1 kg of water at \(100\,^{o}C.\)
Solution.
If t be the final temperature of mixture then, heat lost by 1 kg of water at
\begin{equation*} 100\,^{o}C = 1000\times1\times (100-t) \end{equation*}
and heat gained by 1 kg of water at
\begin{equation*} 0\,^{o}C =1000\times1\times (t-0) \end{equation*}
From the principle of calorimetry,
Heat lost = Heat gained. Therefore,
\begin{equation*} 1000\times1\times (100-t) = 1000\times1\times (t-0) \end{equation*}
\begin{equation*} \therefore\quad t = 50\,^{o}C \end{equation*}
Now increase in entropy of 1kg of water in heating from \(0\,^{o}C\) to \(50\,^{o}C\) is
\begin{equation*} \,dS = \int \frac{\,dQ}{T} \end{equation*}
\begin{equation*} = mc \ln\frac{T_{f}}{T_{i}} = 1000\times 1\times \ln \frac{323}{273} = 168 \,cal/K \end{equation*}
Decrease in entropy of 1kg of water in cooling from \(100\,^{o}C\) to \(50\,^{o}C\) is
\begin{equation*} \,dS = \int \frac{\,dQ}{T} \end{equation*}
\begin{equation*} = mc \ln\frac{T_{f}}{T_{i}} = 1000\times 1\times \ln \frac{323}{373} = -144 \,cal/K \end{equation*}
Therefore the net increase in entropy of the system = 168-144 = 24 cal/K.
PrevTopNext