‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure.  The maximum temperature of the gas during the process will be :

Answers (1)
S Sayak

P as a function of V for the given process would be as follows

\\\frac{P-P_{0}}{V-V_{0}}=\frac{2P_{0}-P_{0}}{V_{0}-2V_{0}}\\ P=-\frac{P_{0}}{V_{0}}V+2P_{0}

Temperature is given as

\\T=\frac{PV}{nR}\\ T=\frac{P_{0}}{nR}\left (- \frac{V^{2}}{V_{0}}+2V\right )

To find the point where temperature would be maximum we see where the slope is 0

\\\frac{dT}{dV}=0\\ \frac{d}{dV}\left ( \frac{P_{0}}{nR}\left (- \frac{V^{2}}{V_{0}}+2V\right ) \right )=0\\ \frac{d}{dV}\left (-\frac{V^{2}}{V_{0}}+2V \right )=0 \\-\frac{2V}{V_{0}}+2=0\\ V=V_{0}

Maximum temperature is therefore at point (V0, 2P0)

T_{max}=\frac{2P_{0}V_{0}}{nR}

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