Question

$2$ moles of $He$ gas $(\gamma =\frac{5}{3})$ are initially at a temperature of ${27}^{\circ }C$ and occupy a volume of $20L$. The gas is first expanded at constant pressure untill the volume is double, then undergoes reversible adiabatic change until the temperature returns to its initial value.
What will be the final volume of gas in liters? (take $2^{1.5}=2.82$)

Answer 1

For state $\left(A\right)P_1=\frac{nRT}{V}=2.46atm$
For isobaric process $\left(AB\right)\frac{V_1}{T_1}=\frac{V_2}{T_2}$
$T_2=600K$
So, temperature and volume after doubling to state $2$ is $40$ L and $600$ K. From here, it goes adiabatic expansion, so that it's final temperature is $300$. Now, the final volume is calculated as follows:
For adiabatic process $\left(BC\right)T{V}^{\gamma -1}=constant$
Where, $\gamma =\frac{C_p}{C_V}$
$600(40{)}^{\gamma -1}=300(V_1{)}^{\gamma -1}$
$\Rightarrow 2=(\frac{V_1}{40}{)}^{\gamma -1}$
$\Rightarrow 2=(\frac{V_1}{40}{)}^{\frac{5}{3}-1}=(\frac{V_1}{40}{)}^{\frac{2}{3}}$
$\Rightarrow V_1=40\times 2^{\frac{3}{2}}=40\times 2.82=113.13L$ please expand the solution. So temperature and volume after doubling to state 2 is 40 and 600. From here it undergoes adiabatic expansion so that its final temperature is 300. so the final volume is calculated as follows please expand the solution. So temperature and volume after doubling to state 2 is 40 and 600. From here it undergoes adiabatic expansion so that its final temperature is 300. so the final volume is calculated as followsplease expand the solution. So temperature and volume after doubling to state 2 is 40 and 600. please expand the solution. So temperature and volume after doubling to state 2 is 40 and 600.