Question

At ${27}^{\circ }C$, three moles of an ideal diatomic gas occupy a volume $V$. If the gas expands adiabatically to a volume $3V$, calculate the work done by the gas during the process.
$(R=8.31\text{J/mol-K},3^{0.4}=1.55)$

• A. $2767\text{J}$
• B. $6669\text{J}$
• C. $6000\text{J}$
• D. $6583\text{J}$

Answer 1

The correct option is B $6669\text{J}$

Given that,
Initial temperature, $T_1={27}^{\circ }C=27+273=300\text{K}$
Initial volume, $V_1=V$
Final volume, $V_2=3V$
$\gamma =1.4$ (for diatomic gas)
For adiabatic process,
$T{V}^{\gamma -1}=\text{constant}$
$T_1{V_1}^{\gamma -1}=T_2{V_2}^{\gamma -1}$
$T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
$\therefore T_2=300{\left(\frac{V}{3V}\right)}^{1.4-1}=193\text{K}$
As we know that, change in internal energy
$\Delta U=n{C}_v\Delta T$
$=3\times \left(\frac{5}{2}R\right)\Delta T$
$=3\times \frac{5}{2}\times 8.31\times (193-300)$
$=-6669\text{J}$
Negative sign means internal energy will decrease.
From first law of thermodynamics,
$\Delta Q=\Delta U+\Delta W$
$\Rightarrow \Delta W=-\Delta U=-(-6669)=6669\text{J}$$(\Delta Q=0$ for adiabatic process$)$

Alternative:
Work done for an adiabatic process, $W=\frac{nR(T_1-T_2)}{\gamma -1}$
$W=\frac{3\times 8.31(300-193)}{1.4-1}\approx 6669\text{J}$

Tips: In a closed adiabatic process, work done is equal to the change in internal energy of the system.