Question

At $27{}^{\circ }C$ two moles of an ideal monatomic gas occupy a volume $V$. The gas expands adiabatically to a volume $2V$.
Given ${\left(\frac{1}{2}\right)}^{\frac{2}{3}}=0.63$.

• A. The final temperatre of gass is $500K$.
• B. Change in internal energy will be $-2767J$.
• C. The work done by the gas during process $2767J$.
• D. The work done by the gas during process $1383.5J$.

Answer 1

Answer: (B), (C)

The correct options are
B Change in internal energy will be $-2767J$.
C The work done by the gas during process $2767J$.

Given, $T_1={27}^{\circ }C=300K$ , $n=2$ moles, $\gamma =5/3$
$V_1=V,V_2=2V,C_V=\frac{3R}{2}$ for monatomic gas

i) To find final temperature: $T_2$
In adiabatic process, $P{V}^{\gamma }$ = constant,
$\therefore T{V}^{\gamma -1}$ = constant,
$P^{1-\gamma }{T}^{\gamma }$=constant.
$\therefore T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}\Rightarrow T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
or $T_2=300{\left(\frac{V}{2V}\right)}^{\frac{5}{3}-1}=300{\left(\frac{1}{2}\right)}^{2/3}=300\times 0.63$
or $T_2=189K$

ii) To find change in internal energy $=\Delta U$.
$\Delta U=n{C}_V\Delta T$
or $\Delta U=3\times 8.31\times (189-300)$
or $\Delta U=-2767J$

iii) Work done
The process is adiabatic, so $\Delta Q=0$
According to first law of thermodynamics
$\Delta Q=\Delta W+\Delta U\therefore 0=\Delta W+(-2767)$
or $\Delta W=2767J$.