Numerical Solutions
Chapter 4: First Law of Thermodynamics
From course: Physics Grade XII
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  1. [2080 ‘P’] A quantity of monoatomic gas at 30oC is compressed suddenly to [latex](\frac{8}{27})^{th}[/latex] parts of its initial volume. Find the change in temperature assuming [latex]\gamma(=\frac{C_p}{C_v})[/latex] to be [latex]\frac{5}{3}[/latex]. [2]
    Solution:
    Initial Case:
    Temperature (T­1) = 30oC = 30 + 273 = 303 K
    Volume (V1) = x
    Final Case:
    Volume (V2) = [latex]\frac{8x}{27}[/latex]
    Change in temperature ([latex]\Delta T)[/latex] =?
    Given, [latex]\gamma = \frac{5}{3}[/latex]
    We know that,
    [latex]T_1V_1^{\gamma – 1} = T_2 V_2^{\gamma – 1}[/latex]

    Or, [latex]T_2 = T_1 \times (\frac{V_1}{V_2})^{\gamma – 1}[/latex]

    Or, [latex]T_2 = 303\times (\frac{x}{\frac{8x}{27}})^{5/3 – 1}[/latex]

    Or, [latex]T_2 = 303\times (\frac{27}{8})^{2/3}[/latex]

    Or, [latex]T_2[/latex] = 681.75 K

    So, change in temperature ([latex]\Delta T)[/latex] = 378.75 K
  1. [2076 ‘B’] A certain volume of dry air at NTP is allowed to expand five times of its original volume under adiabatic condition. Calculate the final pressure and temperature. ([latex]\gamma[/latex] = 1.4).
    Solution:
    Given,
    Initial pressure ([latex]P_1) = 1.01\times 10^5 Nm^{-2}[/latex]
    Initial volume ([latex]V_1)[/latex] = V (let)
    Initial temperature ([latex]T_1)[/latex] = 273 K
    Final volume ([latex]V_2)[/latex] = 5V
    Final pressure ([latex]P_2)[/latex] = ?
    Final temperature ([latex]T_2)[/latex] = ?
    We know that, for adiabatic process,
    [latex]P_1V_1^\gamma = P_2V_2^\gamma[/latex]

    Or, [latex]P_2 = \frac{P_1V_1^\gamma}{V_2^\gamma}[/latex]

    = [latex](1.01\times 10^5)\times (\frac{V}{5V})^{1.4}[/latex]

    = [latex] 1.061\times 10^4 Nm^{-2}[/latex]
    Again, we have,
    [latex]T_1V_1^{\gamma – 1} = T_2V_2^{\gamma – 1}[/latex]

    Or, [latex]T_2 = \frac{T_1V_1^{\gamma – 1}}{V_2^{\gamma – 1}}[/latex]

    = [latex]273 \times (\frac{V}{5V})^{1.4 – 1}[/latex]

    = 143.408 K
  1. [2075 ‘S’] A certain volume of dry air at NTP is allowed to expand four times of its original volume under (i) isothermal conditions (ii) adiabatic conditions. Calculate the final pressure and temperature in each case. ([latex]\gamma = 1.4[/latex])
    Solution:
    Given,
    Initial pressure (P1) = 1.01 x 105 Nm-2
    Initial volume (V1) = V (let)
    Initial temperature (T1) = 273 K
    Final volume (V2) = 4V
    Final pressure (P2) =?
    Final temperature (T2) =?
    (i) For isothermal conditions,
    We know,
    [latex]P_1V_1 = P_2V_2[/latex]

    Or, [latex]P_2 = \frac{P_1V_1}{V_2}[/latex]

    = [latex]\frac{(1.01\times 10^5)\times V}{4V}[/latex]

    = [latex]2.525\times 10^4 Nm^{-2}[/latex]
    Again, for isothermal process, the temperature is constant. So, final temperature (T2) = 273 K.
    (ii) For adiabatic conditions,
    We know,
    [latex]P_1V_1^\gamma = P_2V_2^\gamma[/latex]

    Or, [latex]P_2 = \frac{P_1V_1^\gamma}{V_2^\gamma}[/latex]

    = [latex](1.01\times 10^5)\times (\frac{V}{4V})^{1.4}[/latex]

    = [latex]1.45\times 10^4 Nm^{-2}[/latex]

    Again, we have,
    [latex]T_1V_1^{\gamma – 1} = T_2V_2^{\gamma – 1}[/latex]

    Or, [latex]T_2 = \frac{T_1V_1^{\gamma – 1}}{V_2^{\gamma – 1}}[/latex]

    = [latex]273 \times (\frac{V}{4V})^{1.4 – 1}[/latex]

    = 156.8 K
  1. [2075 ‘B’] A gas in a cylinder initially at a temperature of 10oC and one atmospheric pressure is to be compressed adiabatically to 1/8 of its volume. Find the final temperature. (Take [latex]\gamma[/latex] = 1.4)
    Solution:
    Given,
    Initial volume (V1) = V (let)
    Initial temperature (T1) = 10oC = 10 + 273 = 283 K
    Final temperature (T2) =?
    Now, For adiabatic conditions, We know,
    [latex]T_1V_1^{\gamma – 1} = T_2V_2^{\gamma – 1}[/latex]

    Or, [latex]T_2 = \frac{T_1V_1^{\gamma – 1}}{V_2^{\gamma – 1}}[/latex]

    = [latex]283 \times (\frac{V}{\frac{V}{8}})^{1.4 – 1}[/latex]

    = [latex]283\times (8)^{0.4}[/latex]

    = 650.163 K
  1. [2073 ‘D’ / 2067 ‘S’] A gas in a cylinder is initially at a temperature of 17oC and pressure 1.01 x 105 Nm-2. If it is compressed adiabatically to one-eighth of its original volume, what would be the final temperature and pressure of the gas?
    Solution:
    Given,
    Initial temperature (T1) = 17oC = 17 + 273 = 290 K
    Initial pressure (P1) = 1.01 x 105 Nm-2
    Initial volume (V1) = V (let)
    Final volume (V2) = [latex]\frac{V}{8}[/latex]
    Final temperature (T2) =?
    Final pressure (P2) =?
    We know that, for adiabatic process,
    [latex]T_1V_1^{\gamma – 1} = T_2V_2^{\gamma – 1}[/latex]

    [latex]T_2 = \frac{T_1V_1^{\gamma – 1}}{V_2^{\gamma – 1}}[/latex]

    [latex] = 290\times (\frac{V}{8V})^{1.4 – 1}[/latex]

          = 666.245 K.
    Again, we have,
    [latex]P_1V_1^\gamma = P_2V_2^\gamma[/latex]

     P2 = [latex]\frac{P_1V_1^\gamma}{V_2^\gamma}[/latex]

          = [latex](1.01\times 10^5)\times (\frac{V}{\frac{V}{8}})^{1.4}[/latex]

          = [latex](1.01 \times 10^5 \times (8)^{1.4} [/latex]     

    = 1.856 x 106 Nm-2.
  1. [2072 ‘S’] For hydrogen, the molar heat capacities at constant volume and constant pressure are 20.5 Jmol-1K-1 and 28.8 Jmol-1K-1. Calculate
    i.    The heat needed to raise the temperature of 8g of hydrogen from 10oC to 15oC at constant pressure.     
    ii.   The increase in internal energy of the gas.
    Solution:
    Given,
    Molar heat capacity at constant volume ([latex]C_v[/latex]) = 20.5 Jmol-1K-1
    Molar heat capacity at constant pressure ([latex]C_p[/latex]) = 28.8 Jmol-1K-1
    (i) Mass of hydrogen (m) = 8 g
    Change in temperature ([latex]\Delta \theta[/latex]) = 5 K
    Quantity of heat (dQ) =?
    We know,
    dQ = ncp [latex]\Delta \theta[/latex]

          = [latex]\frac{m}{M}C_p \Delta \theta[/latex]

          = [latex]\frac{8}{2}\times 28.8 \times 5[/latex]

          = 576 J
    (ii) Increase in internal energy of the gas ([latex]\Delta U[/latex]) =?
    We know that,
    [latex]\Delta U = nC_v \Delta \theta[/latex]

     = [latex]\frac{m}{M}C_v \Delta \theta[/latex]

          = [latex]\frac{8}{2}\times 20.5\times 5[/latex]

                = 410 J.
  1. [2072 ‘C’] Five moles of an ideal gas are kept at constant temperature of 53oC while the pressure of the gas is increased from 1.00 atm to 3.00 atm. Calculate work done by the gas.
    Solution:
    Given,
    No. of moles of an ideal gas (n) = 5
    Temperature (T) = 53oC = 53 + 273 = 326 K
    Temperature is Constant i.e. Isothermal process.
    Initial pressure (P1) = 1 atm
    Final pressure (P2) = 3 atm
    Workdone by the gas (W) =?
    We know that,
    Workdone in isothermal process,
    W = [latex]nRTln(\frac{P_1}{P_2})[/latex]

          = [latex]5 \times 8.31 \times 326 \times ln(\frac{1}{3})[/latex]

          = – 1.488 x 104 J.
  1. [2071 ‘C’] Air is compressed adiabatically to half its volume. Calculate the change in its temperature. ([latex]\gamma = 1.42[/latex]).
    Solution:
    Given,
    Initial volume (V1) = V (let)
    Final volume (V2) = V/2
    Change in temperature =?
    We know that, for adiabatic process,
    [latex]T_1V_1^{\gamma – 1} = T_2V_2^{\gamma – 1}[/latex]

    [latex]\frac{T_2}{T_1}      = (\frac{V_1}{V_2})^{\gamma – 1}[/latex]

          = [latex](\frac{V}{V/2})^{1.42 – 1}[/latex]

    = [latex](2)^{0.42}[/latex]
          = 1.338
    Now, change in temperature = [latex]\frac{T_2 – T_1}{T_1}[/latex] x 100%
          = [latex](\frac{T_2}{T_1} – 1)\times 100%[/latex]
          = [latex](1.338 – 1)\times 100%[/latex]
          = 33.8 %.
  1. [2070 ‘S’] A monoatomic ideal gas that is initially at a pressure of 1.50 x 105 pa and has a volume of 0.08 m3 compressed adiabatically to a volume of 0.04 m3. (a) What is the final pressure? (b) How much work is done by the gas? (c) What is the ratio of the final temperature of the gas to its initial temperature?
    Solution:
    Given,
    Initial pressure (P1) = 1.50 x 105 pa
    Initial volume (V1) = 0.08 m3
    Final volume (V2) = 0.04 m3
    For monoatomic gas, [latex]\gamma = 1.67[/latex]
    (a) Final pressure (P2) =?
    We know, for adiabatic process,
    [latex]P_1V_1^\gamma = P_2V_2^\gamma[/latex]

    Or, P2 = [latex]\frac{P_1V_1^\gamma}{V_2^\gamma}[/latex]

          = [latex](1.50\times 10^5)\times (\frac{0.08}{0.04})^{1.67}[/latex]

          = (1.50 x 105) x (2)1.67

          = 4.773 x 105 pa
    (b) Workdone by the gas (W) =?
    We know that, workdone in adiabatic process,
    W = [latex]\frac{1}{\gamma – 1}[P_1V_1 – P_2V_2][/latex]

          = [latex]\frac{1}{1.67 – 1}[(1.5\times 10^5)\times 0.08 – (4.773\times 10^5)\times 0.04][/latex]

          = – 1.059 x 104 J
    (c) Ratio of final temperature to initial temperature =?
    We know that,
    [latex]T_1V_1^{\gamma – 1} = T_2V_2^{\gamma – 1}[/latex]

    Or, [latex]\frac{T_2}{T_1} = (\frac{V_1}{V_2})^{\gamma – 1}[/latex]

          = [latex](\frac{0.08}{0.04})^{1.67 – 1}[/latex]

          = 1.591.
  2. [2069 ‘S’] If the ratio of specific heat capacities of a gas is 1.4 and its density at S.T.P. is 0.09 kg/m3. Calculate the values of specific heat capacities at constant pressure and at constant volume.
    Solution:
    Given,
    Ratio of specific heat capacities of a gas ([latex]\gamma[/latex]) = 1.4
    Density at S.T.P. ([latex]\rho[/latex]) = 0.09 kg/m3
    Specific heat capacity at constant pressure (Cp) =?
    Specific heat capacity at constant volume (Cv) =?
    At S.T.P.,
    Pressure (P) = 1.01 x 105 Nm-2
    Temperature (T) = 273 K
    We know that,
    [latex]\gamma = \frac{C_p}{C_v}[/latex]

    Or, 1.4 = [latex]\frac{C_p}{C_v}[/latex]

    Or, Cp = 1.4Cv …………………. (i)
    Again, for ‘m’ mass of a gas,
    PV = mrT
    Or, r = [latex]\frac{PV}{mT}[/latex]

          = [latex]\frac{P}{\frac{m}{V}T}[/latex]

          = [latex]\frac{P}{\rho T} = \frac{(1.01\times 10^5)}{0.09\times 273}[/latex]

          = 4.11 x 103 JKg-1K-1
    Again,
    Cp – Cv = r
    So, 1.4Cv – Cv = 4.11 x 103

    Or, 0.4Cv = 4.11 x 103

    Or, Cv = [latex]\frac{4.11\times 10^3}{0.4} = 1.028\times 10^4[/latex] Jkg-1K-1

    Now, from eqn. (i), we get,
    Cp = 1.4Cv
          = 1.4 x (1.028 x 104)
          = 1.439 x 104 Jkg-1K-1.
  3. [2068 ‘S’] When 1 gm of water at 100oC and normal pressure becomes 1671 cc of steam at 100oC. Calculate the change in its internal energy. (Latent heat of vaporization of water = 2256 x 103 JKg-1).
    Solution:
    Given,
    Mass of water (mw) = 1 gm = 1 x 10-3 Kg
    Atmospheric pressure (P) = 1.01 x 105 Nm-2
    Since, 1g of water at 100oC occupies 1 cm3 volume,
    So, initial volume (V1) = 1 cm3
    Volume of steam (V2) = 1671 cm3
    Change in volume (dV) = V2 – V1
          = 1671 – 1
          = 1670 cm3 = 1670 x 10-6 m3
    So, workdone by water in converting it into steam (dW) = PdV
          = (1.01 x 105) x (1670 x 10-6)
         = 168.67 J
    Now,
    Quantity of heat supplied to 1g of water to change into steam (dQ) = mLv
          = (1 x 10-3) x (2256 x 103)
          = 2256 J
    So, change in internal energy (dU) = dQ – dW
          = 2256 – 168.67 = 2087.33 J.
  4. [2067, 2066] A litre of air, initially at 20oC and at 760 mm of Hg pressure, is heated at constant pressure until its volume doubled. Find
    (i) the temperature,
    (ii) external work done by the air in expanding, and
    (iii) the quantity of heat supplied. (Specific heat capacity at constant volume = 714 JKg-1K-1, density of air at STP = 1.293 Kgm-3).
    Solution:
    Given,
    Initial temperature (T1) = 20oC = 20 + 273 = 293 K
    Pressure (P) = 760 mm of Hg = 0.76 x 13600 x 9.8 = 1.01 x 105 Nm-2
    Initial volume (V1) = 1 litre = 1 x 10-3 m3
    Final volume (V2) = 2 litre = 2 x 10-3 m3
    Final temperature (T2) =?
    We know that, at constant pressure,
    [latex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/latex]

    Or, [latex]T_2 = \frac{V_2T_1}{V_1} [/latex]

          = [latex]\frac{(2\times 10^{-3})\times 293}{(1\times 10^{-3})}[/latex]

          = 586 K
    We know,
    External workdone (dW) = PdV
          = P(V2 – V1)
          = (1.01 x 105) x (2 x 10-3 – 1 x 10-3)
          = 101 J
    Now, at STP,
    Volume (Vo) =?
    We have,
    [latex]\frac{V_o}{T_o} = \frac{V_1}{T_1}[/latex]

    Or, [latex]\frac{V_o}{273} = \frac{1\times 10^{-3}}{293}[/latex]

    Or, Vo = 9.317 x 10-4 m3
    Now, mass of air (m) = density x volume
         = 1.293 x (9.317 x 10-4)
          = 1.205 x 10-3 kg
    Again, change in internal energy (dU) = mCvdT
          = (1.205 x 10-3) x 714 x (586 – 293)
         = 252.088 J
    So, quantity of heat supplied (dQ) = dU + dW
          = 252.088 + 101
          = 353.088 J.
  5. [2066] An ideal gas initially at 4 atmosphere and 300 K is permitted to expand adiabatically twice its initial volume. Find the final pressure and temperature if the gas is
    i.    Monoatomic Cv = [latex]\frac{3}{2}R[/latex]     
    ii.   Diatomic with Cv = [latex]\frac{5}{2}R[/latex]
    Solution:
    Given,
    Initial pressure (P1) = 4 atm
    Initial temperature (T1) = 300 K
    Initial volume (V1) = V
    Final volume (V2) = 2V
    Final pressure (P2) =?
    Final temperature (T2) =?
    (i) For monoatomic gas, Cv = [latex]\frac{3}{2}R[/latex]

    [latex]\gamma = \frac{C_p}{C_v}[/latex]

          = [latex]\frac{C_v+R}{C_v}[/latex]

          = [latex]1 + \frac{R}{C_v}[/latex]

    = [latex]1 + \frac{R}{\frac{3}{2}R}[/latex]

    = 1 + [latex]\frac{2}{3} = \frac{5}{3}[/latex]

          =  1.67
    We know that,
    [latex]P_1V_1^\gamma = P_2V_2^\gamma[/latex]

    Or, P2 = [latex]\frac{P_1V_1^\gamma}{V_2^\gamma}[/latex]

          = [latex]4\times (\frac{V}{2V})^{1.67}[/latex]

          = 1.257 atm
    Also, [latex]T_1V_1^{\gamma – 1} = T_2V_2^{\gamma – 1}[/latex]

    Or, T2 = [latex]\frac{T_1V_1^{\gamma – 1}}{V_2^{\gamma – 1}} [/latex]

          = [latex]300\times (\frac{V}{2V})^{1.67 – 1}[/latex]

          = 188.552 K
    (ii) For diatomic gas, Cv = [latex]\frac{5}{2}R[/latex]

    [latex]\gamma = \frac{C_p}{C_v}[/latex]

          = [latex]\frac{C_v+R}{C_v}[/latex]

          = [latex]1 + \frac{R}{C_v}[/latex]

    = [latex]1 + \frac{R}{\frac{5}{2}R}[/latex]

    = 1 + [latex]\frac{2}{5} = \frac{7}{5}[/latex]

          =  1.4
    We know that,
    [latex]P_1V_1^\gamma = P_2V_2^\gamma[/latex]

    Or, P2 = [latex]\frac{P_1V_1^\gamma}{V_2^\gamma}[/latex]

          = [latex]4\times (\frac{V}{2V})^{1.4}[/latex]

          = 1.516 atm
    Also, [latex]T_1V_1^{\gamma – 1} = T_2V_2^{\gamma – 1}[/latex]

    Or, T2 = [latex]\frac{T_1V_1^{\gamma – 1}}{V_2^{\gamma – 1}} [/latex]

          = [latex]300\times (\frac{V}{2V})^{1.4 – 1}[/latex]

          = 227.357 K
  6. [2062] An ideal gas is slowly compressed at constant temperature of 50oC to one half of its original volume. In this process, 80 cal of heat was given. How much work was done and what was the change in the internal energy of the gas? Assume one mole of an ideal gas.
    Solution
    Given,
    Temperature (T) = 50oC = 50 + 273 = 323 K
    Initial volume (V1) = V
    Final volume (V2) = V/2|
    Quantity of heat given (dQ) = 80 cal = 80 x 4.2
          = 336 J
    Workdone (W) =?
    Change in internal energy of the gas (dU) =?
    No. of mole (n) = 1
    We know that,
    Workdone in isothermal process (W) = [latex]nRTln\frac{V_2}{V_1}[/latex]

          = 1 x 8.314 x 323 [latex]ln(\frac{V/2}{V})[/latex]

          = – 1861.393 J
    Now, change in internal energy (dU) = dQ – dW
          = 336 – ( – 1861.393)
          = 2197.393 J.

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