Question

An ideal gas has a specific heat at constant pressure to be $C_p=\frac{5}{2}R$. The gas is kept in a closed vessel of volume $0.0083m^3$ at a temperature of $300K$ and a pressure of $1.6\times {10}^{6}N/m^2$. An amount of $2.49\times {10}^{4}J$ of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas. Give $R=8.3J/mo{l}^{-1}{K}^{-1}$.

• A. $T_2=675K$, $P_2=1.8\times {10}^{6}N/m^2$
• B. $T_3=65K$, $P_2=1.8\times {10}^{6}N/m^2$
• C. $T_2=675K$, $P_2=3.6\times {10}^{6}N/m^2$
• D. $T_3=65K$, $P_2=3.6\times {10}^{6}N/m^2$

Answer 1

The correct option is C $T_2=675K$, $P_2=3.6\times {10}^{6}N/m^2$

As here volume of gas remains constant, we have

${\left(\Delta Q\right)}_v=\mu C_v\Delta T$

Here, $C_v=C_p-R=\left(\frac{5}{2}\right)R-R=\left(\frac{3}{2}\right)R$

$\mu =\frac{PV}{RT}=\frac{1.6\times {10}^6\times 8.3\times {10}^{-3}}{8.3\times 300}=\frac{16}{3}$

Now, ${\left(\Delta Q\right)}_v=\mu C_v\Delta T$

${\left(\Delta Q\right)}_v$ $=$$2.49\times {10}^4$ J

$R=8.3$

$2.49\times {10}^4=\frac{16}{3}\times \frac{3}{2}\times 8.3\Delta T$
i.e., $\Delta T=375$ or $T_2-T_1=375$, i.e., $T_2=(375+300)=675K$
Now as for a given mass of an ideal gas at constant volume, $P\alpha T$
$\left(\frac{P_2}{P_1}\right)=\left(\frac{T_2}{T_1}\right)$
$P_2=\frac{675}{300}\times 1.6\times {10}^6=3.6\times {10}^{6}N/m^2$