The change in internal energy of a given mass of gas, when its volume changes from $V$ to 2$V$ at constant pressure $p$ is $(\frac{C_{p}}{C_{v}} = \gamma,$ universal gas constant=R)

Question

Dr. Abubacker · Accepted Answer

(d) In case of gases whatever be the process
$\mathrm{\Delta}U = nC_{V}\mathrm{\Delta}T$
or $\mathrm{\Delta}U = n\mathrm{\Delta}\left( \frac{R}{\gamma - 1} \right)$ -------(i)
$pV = nRT_{1}$
$2pV = nRT_{2}$
$\Rightarrow \ \ \ \ \ \ pV = nR\left( T_{2}{- T}_{1} \right)$
$\frac{pV}{R} = n\mathrm{\Delta}\ T\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (T_{2} - T_{1} < \mathrm{\Delta}T)$
Substituting in Eq. (i)
$\mathrm{\Delta}U = \frac{pV}{(\gamma - 1)}$

Q. The change in internal energy of a given mass of gas, when its volume changes from $V$ to 2$V$ at constant pressure $p$ is $(\frac{C_{p}}{C_{v}} = \gamma,$ universal gas constant=R)

Explanation

Reviewed by Dr. Abubacker

Want more practice?