$n$ moles of an ideal gas undergo process $A B C$ as shown in a figure. The maximum temperature of the gas during the process $A B$ is:

\frac{4 P_{o} V_{o}}{n R}

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\frac{P_{o} V_{o}}{n R}

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\frac{2 P_{o} V_{o}}{n R}

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\frac{3 P_{o} V_{o}}{n R}

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Solution

The correct option is A $\frac{4 P_{o} V_{o}}{n R}$
Relation between pressure and volume along AB is:

$P = P_{o} + \frac{P_{o}}{V_{o}} (V - V_{o})$

According to ideal gas law,

$P V = n R T$

$T = \frac{P V}{n R}$

$T \propto P T \propto V$

So, the temperature will be maximum when pressure and volume will have equal magnitude, which is at the midpoint of the line AB i.e at

$P = 2 P_{o}, V = 2 V_{o}$

So, $T_{m a x} = \frac{4 P_{o} V_{o}}{n R}$

Hence, the correct option is $A$ .

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Similar questions

A \to B

A B C^{'}

A B C^{'}

x (\frac{P V}{n R})

x

\to

P^{6} V^{5} =

n

5 K .

^{'} n^{'}

A \to B

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