The volume $V$ of one mole of a monatomic gas changes with temperature $T$ according to the relation $V = K T^{2 / 3}$ . The work done when temperature changes by $90 K$ will be $x R$ . The value of $x$ is ________.
$(R =$ Universal gas constant $)$

Open in App

Solution

Given:
$V = K T^{2 / 3}$

We know that, work done,

$w = \int P d V = \int \frac{n R T}{V} d V$

$\Rightarrow w = \int \frac{1 \times R T}{K T^{2 / 3}} d (K T^{2 / 3})$

$\Rightarrow w = \frac{2 R}{3} \int d T$

On putting limits, we get,

$w = \frac{2 R}{3} \int_{T_{o}}^{T_{o} + 90} d T$

$\Rightarrow w = \frac{2 R}{3} [(T_{o} + 90) - T_{o}]$

$\Rightarrow w = 60 R = x R$

$\Rightarrow x = 60$

Suggest Corrections

Similar questions

V

T

V = K T^{2 / 3}

90 K

x R

x

(R =

)

V = k T^{\frac{2}{3}}

30^{\circ} C

V = K T^{2 / 3}

V = K T^{2 / 3}

30^{\circ} C

V = k T^{2 / 3}

30^{\circ} C

Join BYJU'S Learning Program