NEET Physics Question - Thermal Physics

The work done by 2 moles of polyatomic gas \( \gamma = \frac{4}{3} \) initially at room temperature to increase its volume eight time during adiabatic process will be (Take \(R = 2\text{ cal mol}^{-1}\text{ K}^{-1}\) and room temperature 27°C)

900 cal

600 cal

1800 cal

1200 cal

Solution:

In adiabatic process, \(T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}\). Here, \(\gamma - 1 = 1/3\). Since \(V_2/V_1 = 8\), we get \(T_2 = T_1 (1/8)^{1/3} = T_1/2 = 300/2 = 150\text{ K}\). The work done is \(W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{2 \times 2 \times (300 - 150)}{1/3} = 1800\text{ cal}\).

Rankers Physics

Work Done in Adiabatic Process

Solution:

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