One mole of an ideal gas undergoes a process P=P01+(3V0/V)2. Here P0 and V0 are constants. Volume is changed from V=V0 to V=3V0

P V = n R T , P = P 0 1 + ( 3 V 0 V ) 2 , a t V = 3 V 0 , P = P 0 2 Equation (i) P 0 2 .3 V 0 = n R T , T = 3 P 0 V 0 2 R When V = V 0 , P = P 0 2 + ( 3 V 0 V 0 ) . = P 0 10 , P 0 10 V 0 = 1 × R T , T = P 0 V 0 10 R

Change in temperature is 5P 0 V 0 8 R

Temperature of gas when its volume is 3 V 0 is 3 P 0 V 0 2 R

Temperature of gas when its volume V 0 is P 0 V 0 10 R

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One mole of an ideal gas undergoes a process $P = \frac{P_{0}}{1 + {(3 V_{0} / V)}^{2}} .$ Here $P_{0}$ and $V_{0}$ are constants. Volume is changed from $V = V_{0} t o V = 3 V_{0}$

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Temperature remains constant

Change in temperature is $\frac{{5P}_{0} V_{0}}{8 R}$

Temperature of gas when its volume is $3 V_{0}$ is $\frac{3 P_{0} V_{0}}{2 R}$

Temperature of gas when its volume $V_{0}$ is $\frac{P_{0} V_{0}}{10 R}$

answer is A, B.

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Detailed Solution

$P V = n R T, P = \frac{P_{0}}{1 + {(\frac{3 V_{0}}{V})}^{2}}, a t V = 3 V_{0}, P = \frac{P_{0}}{2}$

Equation (i) $\frac{P_{0}}{2} .3 V_{0} = n R T, T = \frac{3 P_{0} V_{0}}{2 R}$ When $V = V_{0}, P = \frac{P_{0}}{2 + (\frac{3 V_{0}}{V_{0}})} . = \frac{P_{0}}{10}, \frac{P_{0}}{10} V_{0} = 1 \times R T, T = \frac{P_{0} V_{0}}{10 R}$