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(D) From the graph, the process $1 ightarrow 2$ is an isobaric process because $V \propto T$ (i.e., $V/T = 	ext{constant}$), which implies pressure

Vedclass · Accepted Answer

$\frac{\gamma}{\gamma - 1}$

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(D) From the graph, the process $1 ightarrow 2$ is an isobaric process because $V \propto T$ (i.e., $V/T = 	ext{constant}$), which implies pressure $P$ is constant.For an isobaric process, the heat supplied is given by $\Delta Q = n C_P \Delta T$.The work done by the gas is $\Delta W = P \Delta V = n R \Delta T$.We know that for an ideal gas, $C_P = \frac{\gamma R}{\gamma - 1}$.Substituting this into the expression for $\Delta Q$, we get $\Delta Q = n \left( \frac{\gamma R}{\gamma - 1} ight) \Delta T$.Now, the ratio $\Delta Q : \Delta W$ is:$\frac{\Delta Q}{\Delta W} = \frac{n \left( \frac{\gamma R}{\gamma - 1} ight) \Delta T}{n R \Delta T} = \frac{\gamma}{\gamma - 1}$.

An ideal gas undergoes the process $1 \rightarrow 2$ as shown in the figure. The heat supplied and work done in the process are $\Delta Q$ and $\Delta W$ respectively. The ratio $\Delta Q : \Delta W$ is

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