An ideal gas is expanding such that PT^2 = co | vedclass

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Question

Given process equation is, $\mathrm{pT}^{2}=$ constant Using $\mathrm{PV}=\mathrm{nRT}$

We can have, $\frac{\mathrm{nRT}}{\mathrm{V}} 	imes \mathr

Vedclass · Accepted Answer

$\frac{3}{T}$

Vedclass · Answer

Given process equation is, $\mathrm{pT}^{2}=$ constant Using $\mathrm{PV}=\mathrm{nRT}$

We can have, $\frac{\mathrm{nRT}}{\mathrm{V}} 	imes \mathrm{T}^{2}=$ constant

Or        $\frac{\mathrm{T}^{3}}{\mathrm{V}}=\mathrm{constant}$

By differentiating above equation on w.r. to $v$ $\frac{3 \mathrm{T}^{3} \mathrm{dT}}{\mathrm{V}}-\frac{\mathrm{T}^{3} 	imes \mathrm{dV}}{\mathrm{V}^{2}}=0$

$	ext { or } \gamma=\frac{\mathrm{dV}}{\mathrm{V}} 	imes \frac{1}{\mathrm{dT}}=\frac{3}{\mathrm{T}}$

In above equation left side term is representing coefficient of volume expansion of gas, as from the definition. The coefficient of volume expansion is defined as fractional change in volume per unit rise in temperature.

An ideal gas is expanding such that $PT^2 =$ constant. The coefficient of volume expansion of the gas is

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