Question

The initial pressure, volume and temperature of an ideal gas are $1\text{atm},200\text{L}$ and $300\text{K}$ respectively. The volume and pressure of the ideal gas are adiabatically changed to $74.3\text{L}$ and $4\text{atm}$ respectively. Then:
[Take $R=0.082{\text{L atm K}}^{-1}{\text{mol}}^{-1}$]

• A. Gas is monoatomic
• B. Gas is diatomic
• C. Final temperature is $446\text{K}$
• D. Gas has $8.13\text{moles}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B Gas is diatomic
C Final temperature is $446\text{K}$
D Gas has $8.13\text{moles}$

Given that
Initial volume $\left(V_i\right)=200\text{L}$
Final volume $\left(V_f\right)=74.3\text{L}$
Initial pressure $\left(P_i\right)=1\text{atm}$
Final pressure $\left(P_f\right)=4\text{atm}$
Initial temperature $\left(T_i\right)=300K$
For adiabatic process.
$P_i{V}_i^{\gamma }=P_f{V}_f^{\gamma }$
$\Rightarrow P_f=P_i{\left(\frac{V_i}{V_f}\right)}^{\gamma }$
$\Rightarrow 4\text{atm}=\left(1\text{atm}\right){\left(\frac{200\text{L}}{74.3\text{L}}\right)}^{\gamma }$
$\Rightarrow \gamma =\frac{\ln \left(\frac{P_f}{P_i}\right)}{\ln \left(\frac{V_i}{V_f}\right)}=\frac{\ln \left(\frac{4}{1}\right)}{\ln \left(\frac{200}{74.3}\right)}$
$\Rightarrow \gamma \approx 1.4=\frac{7}{5}$ (gas is diatomic)
Now, $PV=nRT$
$\frac{P_f{V}_f}{P_i{V}_i}=\frac{nR{T}_f}{nR{T}_i}$
$\Rightarrow T_f=\frac{P_f{V}_f}{P_i{V}_i}\times T_i$
$\Rightarrow T_f=\frac{4\times 74.3}{1\times 200}\times 300$
$T_f=445.8\cong 446\text{K}$
and
$n=\frac{P_i{V}_i}{R{T}_i}=\frac{1\times 200}{0.082\times 300}(R=0.082{\text{L atm K}}^{-1}{\text{mol}}^{-1})$
$n=8.13\text{moles}$
Hence, options $\left(b\right),\left(c\right)$ and $\left(d\right)$ are correct.