Discussion Forum

5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be $T_{1}$, the work done in the process is

Answer:

Explanation:

 At STP  ,22.4 L of any gas is 1 mole

 $\therefore$ $5.6 L= \frac{5.6}{22.4}=\frac{1}{4} moles =n$

 In adiabatic process

  $T V ^{\gamma-1}= constant$

 $\therefore$    $T_{2} V_{2}^{\gamma-1}= T_{1} V_{1}^{\gamma-1}$

 or     $T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}$

 $\gamma= \frac{C_{p}}{C_{V}}=\frac{5}{3}$ for monoatomic He gas

 $\therefore$ $T_{2}= T_{1}\left(\frac{5.6}{0.7}\right)^{5/3-1}=4T_{1}$

 $\because$ Further in adiabatic process

 $Q=0$

 $\therefore$   $W+ \triangle U=0$

 or $W= -\triangle U$

 = $-n C_{v} \triangle T$

 =$-n\left(\frac{R}{\gamma-1}\right)(T_{2}-T_{1})$

  =$-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)(4T_{1}-T_{1})$

 = $-\frac{9}{8} RT_{1}$

 $\therefore$ correct option is (a) 

Analysis of Question
(i) From a calculation point of view question is moderately tough.
(ii) Volume of gas is decreasing. Therefore, work done by the gas should be negative.