One mole of an ideal gas undergoes a linear p | vedclass

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(C) The process is represented by a straight line in the $p-V$ diagram.To find the equation of the process,we use the two-point form of the equation o

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$\frac{p_0 V}{R}\left(1-\frac{V}{V_0}\right)$

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(C) The process is represented by a straight line in the $p-V$ diagram.To find the equation of the process,we use the two-point form of the equation of a straight line: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.Here,the two points on the line are $(V_1, p_1) = (0, p_0)$ and $(V_2, p_2) = (V_0, 0)$.Substituting these into the equation,we get:$p - p_0 = \frac{0 - p_0}{V_0 - 0}(V - 0)$$p - p_0 = -\frac{p_0}{V_0} V$$p = p_0 - \frac{p_0}{V_0} V = p_0 \left(1 - \frac{V}{V_0}\right)$.For one mole of an ideal gas,the ideal gas equation is $pV = RT$,which implies $p = \frac{RT}{V}$.Substituting the expression for $p$ into the ideal gas equation:$\frac{RT}{V} = p_0 \left(1 - \frac{V}{V_0}\right)$$T = \frac{p_0 V}{R} \left(1 - \frac{V}{V_0}\right)$.

One mole of an ideal gas undergoes a linear process as shown in the figure below. Its temperature expressed as a function of volume $V$ is

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