The P-V graph for a process is shown from 1 t | vedclass

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(C) For an ideal gas,the equation of state is $PV = nRT$,which implies $T = \frac{PV}{nR}$.Along the line segment from $1$ to $2$,the product $PV$ cha

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Heating initially and cooling finally.

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(C) For an ideal gas,the equation of state is $PV = nRT$,which implies $T = \frac{PV}{nR}$.Along the line segment from $1$ to $2$,the product $PV$ changes.Let the line be $P = -mV + c$,where $m$ is the slope and $c$ is the intercept.Then $T(V) = \frac{(-mV + c)V}{nR} = \frac{1}{nR}(-mV^2 + cV)$.This is a downward-opening parabola with respect to $V$.The maximum temperature occurs at the midpoint of the line segment $1-2$.As we move from $1$ to the midpoint,the product $PV$ increases,so the gas is heated.As we move from the midpoint to $2$,the product $PV$ decreases,so the gas is cooled.Thus,the gas is heated initially and cooled finally.

The $P-V$ graph for a process is shown from $1$ to $2$. As $T$ changes,the process for the gas is:

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