The volume ($V$) of a monatomic gas varies with its temperature ($T$), as shown in the graph. The ratio of wor

Question

The volume ($V$) of a monatomic gas varies with its temperature ($T$), as shown in the graph. The ratio of work done by the gas to the heat absorbed by it when it undergoes a change from state $A$ to state $B$ will be:

Accepted Answer

From the description, the graph of Volume ($V$) versus Temperature ($T$) is a straight line passing through the origin (implied by the variation $V \propto T$). This indicates that the process is **isobaric** (constant pressure).

1.  **Work Done ($W$):** In an isobaric process, the work done by the gas is given by:
    $W = P\Delta V = nR\Delta T$

2.  **Heat Absorbed ($Q$):** The heat absorbed at constant pressure is given by:
    $Q = nC_p\Delta T$
    Where $C_p$ is the molar heat capacity at constant pressure.

3.  **Specific Heat for Monatomic Gas:** For a monatomic ideal gas, the molar heat capacity at constant pressure is:
    $C_p = \frac{5}{2}R$ (Source: Appendix, Molar Heat Capacities )

4.  **Calculate Ratio:**
    $	ext{Ratio} = \frac{	ext{Work Done}}{	ext{Heat Absorbed}} = \frac{W}{Q} = \frac{nR\Delta T}{nC_p\Delta T} = \frac{R}{C_p}$
    Substituting $C_p = \frac{5}{2}R$:
    $	ext{Ratio} = \frac{R}{\frac{5}{2}R} = \frac{2}{5}$

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