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Question

Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure.

$w_{1},w_{2},w_{3}$ and $w_{4}$ represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; $\Delta U_{2}$ and $\Delta U_{4}$ are changes in the internal energy for the processes 2 and 4, respectively.

[use $R = 2calK^{- 1}{\text{ }mol}^{- 1}$ ]

The correct option is

Options

A.$w_{1} + w_{2} = 2{\text{ }T}_{1}ln\frac{{\text{ }V}_{2}}{{\text{ }V}_{1}}$
B.$w_{1} + w_{2} + w_{3} + w_{4} = 0$
C.$w_{1} + w_{3} = - 2{\text{ }T}_{1}ln\frac{{\text{ }V}_{2}}{{\text{ }V}_{1}} - 2{\text{ }T}_{2}ln\frac{{\text{ }V}_{4}}{{\text{ }V}_{3}}$
D.$w_{2} + w_{4} = \Delta U_{2} - \Delta U_{4}$

Solution

Sol. [Key Concept: Reversible processes of an ideal gas]

$w_{1} \rightarrow$ isothermal reversible process

$$\begin{matrix} & w_{1} = - nRT_{1}ln\frac{V_{2}}{V_{1}} \\ & w_{1} = - 1 \times R \times T_{1}ln\frac{V_{2}}{V_{1}} \\ & w_{3} = - n \times R \times T_{2}ln\frac{V_{4}}{V_{3}} \end{matrix}$$

$$\begin{matrix} & w_{3} = - 1 \times R \times T_{2}ln\frac{{\text{ }V}_{4}}{{\text{ }V}_{3}} \\ & w_{1} + w_{3} = - {nRT}_{1}ln\left( \frac{{\text{ }V}_{2}}{{\text{ }V}_{1}} \right) - {nRT}_{2}ln\left( \frac{{\text{ }V}_{4}}{{\text{ }V}_{3}} \right) \\ & w_{1} + w_{3} = - 2{\text{ }T}_{1}ln\left( \frac{{\text{ }V}_{2}}{{\text{ }V}_{1}} \right) - 2{\text{ }T}_{2}ln\left( \frac{{\text{ }V}_{4}}{{\text{ }V}_{3}} \right) \end{matrix}$$

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