Question

Two moles of helium gas $(\gamma =5/3)$ are initially at a temperature of ${27}^o$C and occupy a volume of $20$ litres. The gas is first expanded at constant pressure until the volume is doubled. It then undergoes an adiabatic change until the temperature returns to its initial value.
What are the final volume and pressure of the gas?

• A. $P_C=1.44\times {10}^{5}N/m^2,V_C=113.1$ litre.
• B. $P_C=0.44\times {10}^{5}N/m^2,V_C=113.1$ litre.
• C. $P_C=0.44\times {10}^{5}N/m^2,V_C=213.1$ litre.
• D. $P_C=0.44\times {10}^{6}N/m^2,V_C=113.1$ litre.

Answer 1

The correct option is B $P_C=0.44\times {10}^{5}N/m^2,V_C=113.1$ litre.

No of moles $=2$

$r=\frac{5}{3}$

$T_2=300K(27+273=300k)$

$V=20ltr$

$V_f=40ltr$

in constant pressure $\to \frac{V}{T_i}=\frac{V_{f^1}}{T_{f^1}}\Rightarrow T_{f^1}=600k$

in adiabatic process

$T{V}^{r-1}=k$

$T_{f^1}{V}_{f^1}=T_i(V_f{)}^{r-1}$

$\left(600\right)(40{)}^{2/3}=(300\left)\right(V_f{)}^{2/3}$

$\left(2^{3/2}\right)\left(40\right)=V_f$

Solve & we get

$V_f=113.1ltr$

for $P_f$ use eqn of state

$nR=\frac{P_i{V}_i}{T_i}=\frac{P_f{V}_f}{T_f}$ $(T_f=T_i)$

$P_i\left(20\right)=\left(113.1\right)P_f$

$2\times \left(23\right)\times \left(300\right)=P_f\left(\frac{\left(113.1\right)}{1000}\right)$ $1ltr=\frac{m^3}{1000}R=8.3J{k}^{-1}mo{l}^{-1}$

$44,031.8302=P_f$

$P_f=0.44\times {10}^{5}N/m^2$