Question

A container capable of handling a maximum pressure of 2 atm is being filled with helium gas at room temperature (of $27{}^{\circ }$C). What will be the resulting temperature of helium gas, if it suddenly bursts?

• A. $-60{}^{\circ }$C
• B. $-33{}^{\circ }$C
• C. $-30.15{}^{\circ }$C
• D. $-44{}^{\circ }$C

Answer 1

The correct option is B $-33{}^{\circ }$C

Sudden bursting is an adiabatic irreversible process for which,
$q=0$
So, from first law of thermodynamics,
$\Delta U=w$
$n{C}_v\Delta T=-P_{ext}\Delta V$
$n{C}_v(T_2-T_1)=1\times \left[\right(\frac{nR{T}_1}{P_1})-(\frac{nR{T}_2}{P_2}\left)\right]$
$n\times \frac{3R}{2}(T_2-300)=nR\left[\right(\frac{300}{2})-(\frac{T_2}{1}\left)\right]$
$\frac{3}{2}(T_2-300)=[150-T_2]$
$T_2=240K=-{33}^{0}C$