Question

One mole of a gas expands obeying the relation as shown in P-V diagram. The maximum temperature in this process is equal to :

• A. $\frac{P_{0}V_{0}}{R}$
• B. $\frac{3P_{0}V_{0}}{R}$
• C. $\frac{9P_{0}V_{0}}{8R}$
• D. None

Answer 1

Answer: (C)

$\frac{9P_{0}V_{0}}{8R}$

Answer 2

The slope of the graph is, m = –$\frac{p_{0}}{2V_{0}}$

The equation of this graph is given by

P = –$\frac{p_{0}V}{2V_{0}}$ +$\frac{3p_{0}}{2}$

̃ =$\frac{nRT}{V}$= –$\frac{p_{0}}{2V_{0}}$+$\frac{3p_{0}}{2}$

̃ T –$\frac{p_{0}}{2nRV_{0}}$V² +$\frac{3p_{0}}{2nR}$V

$\frac{dT}{dV}$= –$\frac{2p_{0}V}{2nRV_{0}}$+$\frac{3p_{0}}{2nR}$ = 0

̃ =$\frac{p_{0}V}{nRV_{0}}$=$\frac{3P_{0}}{2nR}$ ̃ V =$\frac{3}{2}$V₀

At V =$\frac{3V_{0}}{2}$, temperature will be maximum [n = 1]

T_max = –$\frac{p_{0}}{2RV_{0}}$×$\frac{9V_{0}^{2}}{4}$ +$\frac{3p_{0}}{2R}$×$\frac{3V_{0}}{2}$

̃ –$\frac { 9 } { 8 }$ $\frac{p_{0}V_{0}}{R}$ + $\frac { 9 } { 4 }$ $\frac{p_{0}V_{0}}{R}$

̃ $\frac { 9 } { 8 }$ $\frac{p_{0}V_{0}}{R}$