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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Specific Heat Capacities of Gases

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Specific Heat Capacities of Gases

Half mole of an ideal gas ( γ = 5/3 ) is taken through the cycle...

Half mole of an ideal gas (γ=5/3\gamma = 5/3) is taken through the cycle abcdaabcda as shown in the figure. Take R=253 J K−1 mol−1R = \dfrac{25}{3} \text{ J K}^{-1} \text{ mol}^{-1}. Determine the temperature of the gas in the states aa, bb, cc and dd.
Half mole of an ideal gas ( γ = 5/3 ) is taken through the cycle abcda as shown in the figure. Take R = (25)/(3) J K^

Options

  1. A120 K120 \text{ K}, 240 K240 \text{ K}, 480 K480 \text{ K}, 240 K240 \text{ K}
  2. B60 K60 \text{ K}, 120 K120 \text{ K}, 240 K240 \text{ K}, 120 K120 \text{ K}
  3. C240 K240 \text{ K}, 480 K480 \text{ K}, 960 K960 \text{ K}, 480 K480 \text{ K}
  4. D120 K120 \text{ K}, 120 K120 \text{ K}, 240 K240 \text{ K}, 240 K240 \text{ K}

Correct answer

A. 120 K120 \text{ K}, 240 K240 \text{ K}, 480 K480 \text{ K}, 240 K240 \text{ K}

Step-by-step solution

Using the ideal gas equation PV=nRTPV = nRT, the temperature at any state is given by T=PVnRT = \dfrac{PV}{nR}.

Given:
n=0.5 moln = 0.5 \text{ mol}
R=253 J K−1 mol−1R = \dfrac{25}{3} \text{ J K}^{-1} \text{ mol}^{-1}
nR=0.5×253=256 J K−1nR = 0.5 \times \dfrac{25}{3} = \dfrac{25}{6} \text{ J K}^{-1}

For state aa:
Pa=100 kPa=105 PaP_a = 100 \text{ kPa} = 10^5 \text{ Pa}
Va=5000 cm3=5×10−3 m3V_a = 5000 \text{ cm}^3 = 5 \times 10^{-3} \text{ m}^3
Ta=PaVanR=105×5×10−3256=500×625=120 KT_a = \dfrac{P_a V_a}{nR} = \dfrac{10^5 \times 5 \times 10^{-3}}{\dfrac{25}{6}} = \dfrac{500 \times 6}{25} = 120 \text{ K}

For state bb:
Pb=100 kPa=105 PaP_b = 100 \text{ kPa} = 10^5 \text{ Pa}
Vb=10000 cm3=10×10−3 m3V_b = 10000 \text{ cm}^3 = 10 \times 10^{-3} \text{ m}^3
Tb=PbVbnR=105×10×10−3256=1000×625=240 KT_b = \dfrac{P_b V_b}{nR} = \dfrac{10^5 \times 10 \times 10^{-3}}{\dfrac{25}{6}} = \dfrac{1000 \times 6}{25} = 240 \text{ K}

For state cc:
Pc=200 kPa=2×105 PaP_c = 200 \text{ kPa} = 2 \times 10^5 \text{ Pa}
Vc=10000 cm3=10×10−3 m3V_c = 10000 \text{ cm}^3 = 10 \times 10^{-3} \text{ m}^3
Tc=PcVcnR=2×105×10×10−3256=2000×625=480 KT_c = \dfrac{P_c V_c}{nR} = \dfrac{2 \times 10^5 \times 10 \times 10^{-3}}{\dfrac{25}{6}} = \dfrac{2000 \times 6}{25} = 480 \text{ K}

For state dd:
Pd=200 kPa=2×105 PaP_d = 200 \text{ kPa} = 2 \times 10^5 \text{ Pa}
Vd=5000 cm3=5×10−3 m3V_d = 5000 \text{ cm}^3 = 5 \times 10^{-3} \text{ m}^3
Td=PdVdnR=2×105×5×10−3256=1000×625=240 KT_d = \dfrac{P_d V_d}{nR} = \dfrac{2 \times 10^5 \times 5 \times 10^{-3}}{\dfrac{25}{6}} = \dfrac{1000 \times 6}{25} = 240 \text{ K}
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