Question

A sample of 2 kg monoatomic helium (assumed ideal) is taken
through the process ABC and another sample of 2 kg of the
same gas is taken through the process ADC (see fig). Given
molecular mass of helium = 4.
(a) What is the temperature of helium in each of the states
A, B,C and D?
(b) Is there any way of telling afterwards which sample of
helium went through the process ABC and which went
through the process ADC? Write Yes or No.
(c) How much is the heat involved in the process ABC and
ADC?

Answer 1

Number of gram moles of He
$ \, \, \, \, \, \, \, \, \, \, \, \, \, n=\frac{m}{M}=\frac{2 \times 10^3}{4}=500$
(a)$V_A =10m^3, p_A =5 \times 10^4 N/m^2$
$\therefore \, \, \, \, \, \, T_A=\frac{p_AV_A}{nR}=\frac{(10)(5 \times 10^4)}{(500)(8.31)} K$
or $T_A=120.34 K$
Similarly, $V_B = 10 m^3, p B = 10 \times 10^4 N/m^2 $
$\therefore \, \, \, \, \, T_B=\frac{(10)(10 \times 10^4)}{(500)(8.31)}K$
$\therefore \, \, \, \, \, T_B =240.68 K$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, V_C=20m^3.pC=10 \times 10^4 n/m^2$
$\therefore \, \, \, \, \, T_C=\frac{(20)(10 \times 10^4)}{(500)(8.31)}K$
and $V_D =20 m^3, p_{\upsilon}=5 \times 10^4 N/m^2$
$\therefore \, \, \, \, \, \, \, V_D =\frac{(20)(5 \times 10^4)}{(500)(8.31)}K$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, T_D =240.68 K $
(b) No, it is not possible to tell afterwards which sample
went through the process ABC or ADC. But during the
process if we note down the work done in both the
processes, then the process which require more work
goes through process ABC.
(c) In the process ABC
$\Delta U =nC_V \, \Delta T =n\bigg(\frac{3}{2}R\bigg)(T_C-T_A)$
$ \, \, \, \, \, \, \, =(500)\bigg(\frac{3}{2}\bigg)(8.31)(481.36-120.34)J$
$\Delta U =2.25 \times 10^6 J$
and $\Delta W =$Area under BC
$ \, \, \, \, \, =(20-10)(10) \times 10^4 J=10^6 J$
$\therefore \, \, \, \, \Delta Q_{ABC} =\Delta U + \Delta W =(2.25 \times 10^6 + 10^6)J$
$\Delta Q_{ABC}=3.25 \times 10^6 J$
In the process A D C A U will be same (because it depends on
initial and final temperatures only)
$\Delta W =Area \, under \, AD$
$ =(20-10)(5 \times 10^4)j=0.5 \times 10^6 J$
$\Delta Q_{ADC}=\Delta U+\Delta W =(2.25 \times 10^6 + 0.5 \times 10^6) J$
$\Delta Q_{ADC} =2.75 \times 10^6 J$