Question

$1.6\text{L}$ of helium gas at STP is adiabatically compressed to $0.2\text{L}$. Taking the initial temperature to be ${27}^{\circ }\text{C}$, the work done in the process is
[Take $R=8.314\text{J/mol K}$]

• A. $1600\text{J}$
• B. $400\text{J}$
• C. $800\text{J}$
• D. $200\text{J}$

Answer 1

The correct option is C $800\text{J}$

Given that,

Initial volume, $V_1=1.6\text{L}$
${\gamma }_{He}$ = $\frac{5}{3}$ for monoatomic gas
Final volume $V_2=0.2\text{L}$
Initial temperature $T_1={27}^{\circ }\text{C}=300\text{K}$
Initial pressure $P_1=1\text{atm}$ [STP}

Number of moles of gas is
$n=\frac{\text{Volume of gas at STP}}{22.4\text{litres}}=\frac{1.6}{22.4}=\frac{1}{14}$

For adiabatic compression.
$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
$\Rightarrow T_2=T_1{\left(\frac{V_1}{V_2}\right)}^{\gamma -1}=300{\left(\frac{1.6}{0.2}\right)}^{\frac{5}{3}-1}=300\times 4=1200\text{K}$

We know that work done in an adiabatic process is
$W=\frac{nR\Delta T}{\gamma -1}=\frac{1}{14}\times \frac{8.314\times (1200-300)}{\frac{5}{3}-1}$
$=801.7\approx 800\text{J}$