1 , text{mole} of gas expands with temperatur | vedclass

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Question

(B) Given the relation $V = KT^{2/3}$.
Using the ideal gas law $PV = nRT$, we have $T = \frac{PV}{nR}$.
Substituting $T$ into the given relation: $V

Vedclass · Accepted Answer

$20$

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(B) Given the relation $V = KT^{2/3}$.
Using the ideal gas law $PV = nRT$, we have $T = \frac{PV}{nR}$.
Substituting $T$ into the given relation: $V = K \left( \frac{PV}{nR} ight)^{2/3}$.
Raising both sides to the power of $3/2$: $V^{3/2} = K^{3/2} \frac{PV}{nR}$.
Rearranging the terms: $V^{3/2} / V = \frac{K^{3/2}}{nR} P \Rightarrow V^{1/2} = \frac{K^{3/2}}{nR} P$.
This can be written as $P V^{-1/2} = 	ext{constant}$.
Comparing this with the polytropic process equation $PV^x = 	ext{constant}$, we get $x = -1/2$.
The work done in a polytropic process is given by $W = \frac{nR \Delta T}{1-x}$.
Substituting the values $n = 1$, $\Delta T = 30$, and $x = -1/2$:
$W = \frac{1 	imes R 	imes 30}{1 - (-1/2)} = \frac{30R}{3/2} = 20R$.

$1 \, \text{mole}$ of gas expands with temperature according to the relation $V = KT^{2/3}$. When the temperature changes by $30 \, ^\circ\text{C}$, the work done will be: (in $, R$)

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