Question

The volume of one mole of an ideal with the adiabatic exponent $\gamma$ is varied according to the $V=\frac{a}{T}$. Where $a$ is a gas constant. Find the amount of $h$ obtained by the gas in this process if the gas temperature increased by $△T$.

• A. $nR△T\left[\frac{3-\gamma }{\gamma -1}\right]$.
• B. $nR△T\left[\frac{2-\gamma }{\gamma -1}\right]$.
• C. $nR△T\left[\frac{4-\gamma }{\gamma -1}\right]$.
• D. $nR△T\left[\frac{5-\gamma }{\gamma -1}\right]$.

Answer 1

Answer: (B)

$nR△T\left[\frac{2-\gamma }{\gamma -1}\right]$.

$\text{Given:-}$ Adiabatic exponent $=\gamma$

Volume of one mole of an ideal gas,$V=a/T$

$\text{Solution:-}$

The differential work done by the gas

$dA=PdV=\frac{vR{T}^2}{a}(-\frac{a}{T^2})dT=-vRdT$

$(asPV=vRTandV=\frac{a}{T})$

So, $A=-{\int }_T^{T+\Delta T}vRdT=-vR\Delta T$

From the first law of thermodynamics

$Q=\Delta U+A=\frac{vR}{\gamma -1}\Delta T-vR\Delta T$

$=vR\Delta T\cdot \frac{2-\gamma }{\gamma -1}=R\Delta T\cdot \frac{2-\gamma }{\gamma -1}$

$(Forv=1mole)$

$\text{Hence the correct option is B}$