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Question: n moles of an ideal gas undergoes a process A to B as shown. Maximum temperature of gas during the p...

n moles of an ideal gas undergoes a process A to B as shown. Maximum temperature of gas during the process is –

A

3P0 V0nR\frac { 3 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { \mathrm { nR } }

B

4P0 V0nR\frac { 4 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { \mathrm { nR } }

C

6P0 V0nR\frac { 6 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { \mathrm { nR } }

D

9P0 V0nR\frac { 9 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { \mathrm { nR } }

Answer

4P0 V0nR\frac { 4 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { \mathrm { nR } }

Explanation

Solution

Equation of straight line passing through points A and B is

P – P0 =(V – 3V0)

P = P0 + (2P0−2 V0)\left( \frac { 2 \mathrm { P } _ { 0 } } { - 2 \mathrm {~V} _ { 0 } } \right) (V – 3V0) = P0 –(V – 3V0)

P = P0 V0−P0 V+3P0 V0 V0\frac { \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } - \mathrm { P } _ { 0 } \mathrm {~V} + 3 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { \mathrm {~V} _ { 0 } } = 4P0 V0−P0 V V0\frac { 4 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } - \mathrm { P } _ { 0 } \mathrm {~V} } { \mathrm {~V} _ { 0 } }

Since PV = nR or T = PVnR\frac { \mathrm { PV } } { \mathrm { nR } }

\ T = (4P0 V0−P0 V V0)\left( \frac { 4 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } - \mathrm { P } _ { 0 } \mathrm {~V} } { \mathrm {~V} _ { 0 } } \right) = 4P0 V0 V−P0 V2nRV0\frac { 4 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } \mathrm {~V} - \mathrm { P } _ { 0 } \mathrm {~V} ^ { 2 } } { \mathrm { nR } \mathrm { V } _ { 0 } }

For temperature to be max.

= 0.

or 4P0V0 – 2P0V = 0

or V = 2V0

Hence Tmax. =4P0 V0(2 V0)−P0(2 V0)2 V0nR\frac { 4 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } \left( 2 \mathrm {~V} _ { 0 } \right) - \mathrm { P } _ { 0 } \left( 2 \mathrm {~V} _ { 0 } \right) ^ { 2 } } { \mathrm {~V} _ { 0 } \mathrm { nR } }Tmax.

=8P0 V02−4P0 V02 V0nR\frac { 8 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } ^ { 2 } - 4 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } ^ { 2 } } { \mathrm {~V} _ { 0 } \mathrm { nR } }

Tmax. =4P0 V0nR\frac { 4 \mathrm { P } _ { 0 } \mathrm {~V} _ { 0 } } { \mathrm { nR } }