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Question: The pressure of one mole of ideal gas varies according to the law P = P<sub>0</sub> – aV<sup>2</sup>...

The pressure of one mole of ideal gas varies according to the law P = P0 – aV2 where P0 &a are positive constants. The highest temperature that gas may attain :

A

2P03R(P03α)1/2\frac { 2 \mathrm { P } _ { 0 } } { 3 \mathrm { R } } \left( \frac { \mathrm { P } _ { 0 } } { 3 \alpha } \right) ^ { 1 / 2 }

B

3P02R(P03α)1/2\frac { 3 \mathrm { P } _ { 0 } } { 2 \mathrm { R } } \left( \frac { \mathrm { P } _ { 0 } } { 3 \alpha } \right) ^ { 1 / 2 }

C

P0R(P03α)1/2\frac { \mathrm { P } _ { 0 } } { \mathrm { R } } \left( \frac { \mathrm { P } _ { 0 } } { 3 \alpha } \right) ^ { 1 / 2 }

D

P0R(P0α)1/2\frac { \mathrm { P } _ { 0 } } { \mathrm { R } } \left( \frac { \mathrm { P } _ { 0 } } { \alpha } \right) ^ { 1 / 2 }

Answer

2P03R(P03α)1/2\frac { 2 \mathrm { P } _ { 0 } } { 3 \mathrm { R } } \left( \frac { \mathrm { P } _ { 0 } } { 3 \alpha } \right) ^ { 1 / 2 }

Explanation

Solution

P = P0 – aV2

PV = RT

= P0 – aV2

T = – αV3R\frac { \alpha \mathrm { V } ^ { 3 } } { \mathrm { R } } = 0

P0R\frac { \mathrm { P } _ { 0 } } { \mathrm { R } } – = 0

V = P03α\sqrt { \frac { \mathrm { P } _ { 0 } } { 3 \alpha } }

Now put V in T