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Physics Question on Thermodynamics

′n′'n' moles of an ideal gas undergoes a process A→BA \to B as shown in the figure. The maximum temperature of the gas during the process will be :

A

9 P0 V04nR\frac{9\, P_{0}\, V_{0}}{4nR}

B

3 P0 V02nR\frac{3\, P_{0} \,V_{0}}{2 nR}

C

9 P0 V02nR\frac{9\, P_{0} \,V_{0}}{2 nR}

D

9 P0 V0nR\frac{9\, P_{0} \,V_{0}}{nR}

Answer

9 P0 V04nR\frac{9\, P_{0}\, V_{0}}{4nR}

Explanation

Solution

Equation of line ABAB
y−y1=y2−y1x2−x1(x−x1)y -y_{1} = \frac{y_{2} -y_{1}}{x_{2} -x_{1}}\left(x-x_{1}\right)
P−P0=2P0−P0V0−2V0P - P_{0} = \frac{2P_{0} - P_{0}}{V_{0} - 2V_{0}}
(V−2V0)=−P0V0(V−2V0)\left(V - 2V_{0}\right) = - \frac{P_{0}}{V_{0}} \left(V - 2V_{0}\right)
P=−P0V0V+3P0P = - \frac{P_{0}}{V_{0}} V + 3P_{0}
PV=−P0V0V2+3P0VPV = - \frac{P_{0}}{V_{0}} V^{2} + 3P_{0}V
nRT=−P0V0V2+3P0VnRT = - \frac{P_{0}}{V_{0}}V^{2} + 3P_{0}V
T=1nR(P0V0V2+3P0V)T = \frac{1}{nR}\left(\frac{P_{0}}{V_{0} }V^{2} + 3P_{0}V\right)
dTdV=0\frac{dT}{dV} = 0 (For maximum temperature)
P0V02V+3P0=0\frac{P_{0}}{V_{0}} 2V + 3P_{0} = 0
P0V02V=−3P0\frac{P_{0}}{V_{0}}2V = - 3 P_{0}
V=32V0V = \frac{3}{2}V_{0} (Condition for maximum temperature)
Tmax=1nR(−P0V0×94V02+3P0×32V0)T_{max } = \frac{1}{nR} \left( - \frac{P_{0}}{V_{0} } \times \frac{9}{4} V_{0}^{2}+3P_{0} \times\frac{3}{2}V_{0}\right)
=1nR(−94P0V0+92P0V00)=94P0V0nR= \frac{1}{nR} \left( - \frac{9}{4} P_{0}V_{0} + \frac{9}{2} P_{0}V_{0}0\right) = \frac{9}{4} \frac{P_{0}V_{0}}{nR} 0