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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

When an ideal gas is supplied with 50 J of heat, it expands from...

When an ideal gas is supplied with 50 J50 \text{ J} of heat, it expands from 100 cm3100 \text{ cm}^3 to 200 cm3200 \text{ cm}^3 at a constant pressure of 2.0×105 Pa2.0 \times 10^5 \text{ Pa}. If the initial temperature is 300 K300 \text{ K}, determine the molar heat capacity CpC_p at constant pressure.

Options

  1. A20.8 J K−1 mol−120.8 \text{ J K}^{-1} \text{ mol}^{-1}
  2. B29.1 J K−1 mol−129.1 \text{ J K}^{-1} \text{ mol}^{-1}
  3. C8.31 J K−1 mol−18.31 \text{ J K}^{-1} \text{ mol}^{-1}
  4. D12.5 J K−1 mol−112.5 \text{ J K}^{-1} \text{ mol}^{-1}

Correct answer

A. 20.8 J K−1 mol−120.8 \text{ J K}^{-1} \text{ mol}^{-1}

Step-by-step solution

Given Q=50 JQ = 50 \text{ J}, P=2.0×105 PaP = 2.0 \times 10^5 \text{ Pa}, V1=100 cm3=10−4 m3V_1 = 100 \text{ cm}^3 = 10^{-4} \text{ m}^3, V2=200 cm3=2×10−4 m3V_2 = 200 \text{ cm}^3 = 2 \times 10^{-4} \text{ m}^3, T1=300 KT_1 = 300 \text{ K}.

From the ideal gas equation PV1=nRT1P V_1 = n R T_1, we get:

nR=PV1T1=2.0×105×10−4300=20300=115 J K−1n R = \dfrac{P V_1}{T_1} = \dfrac{2.0 \times 10^5 \times 10^{-4}}{300} = \dfrac{20}{300} = \dfrac{1}{15} \text{ J K}^{-1}

At constant pressure, V∝TV \propto T, so:

V1T1=V2T2⇒100300=200T2⇒T2=600 K\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \Rightarrow \dfrac{100}{300} = \dfrac{200}{T_2} \Rightarrow T_2 = 600 \text{ K}

The change in temperature is ΔT=600−300=300 K\Delta T = 600 - 300 = 300 \text{ K}.

The heat supplied at constant pressure is given by Q=nCpΔTQ = n C_p \Delta T.

Substituting the values:

50=nCp(300)50 = n C_p (300)

Cp=50300n=50300×RnRC_p = \dfrac{50}{300 n} = \dfrac{50}{300} \times \dfrac{R}{n R}

Cp=16×R1/15=156R=2.5RC_p = \dfrac{1}{6} \times \dfrac{R}{1/15} = \dfrac{15}{6} R = 2.5 R

Using R=8.31 J K−1 mol−1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}:

Cp=2.5×8.31=20.775 J K−1 mol−1≈20.8 J K−1 mol−1C_p = 2.5 \times 8.31 = 20.775 \text{ J K}^{-1} \text{ mol}^{-1} \approx 20.8 \text{ J K}^{-1} \text{ mol}^{-1}
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