Heat And Thermodynamics Ques 12
- An ideal gas has a specific heat at constant pressure $C_p=\frac{5 R}{2}$. The gas is kept in a closed vessel of volume $0.0083 \mathrm{~m}^3$, at a temperature of $300 \mathrm{~K}$ and a pressure of $1.6 \times 10^6 \mathrm{~N} / \mathrm{m}^2$. An amount of $2.49 \times 10^4 \mathrm{~J}$ of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas.
(1987, 7M)
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Answer:
Correct Answer: 12.$675 \mathrm{~K}, 3.6 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
Solution:
Formula:
- Vessel is closed. Therefore, $\Delta W=0$
or
$ \begin{aligned} \Delta Q & =\Delta U=n C_V \Delta T \\ \Delta T & =\frac{\Delta Q}{n C_V}=\frac{\Delta Q}{\left(\frac{p V}{R T}\right)\left(C_p-R\right)} \\ & =\frac{(\Delta Q)(R T)}{p V\left(C_p-R\right)} \end{aligned} $
Substituting the values, we have
$ \begin{aligned} \Delta T & =\frac{\left(2.49 \times 10^4\right)(300)}{\left(1.6 \times 10^6\right)(0.0083)(3 / 2)}=375 \mathrm{~K} \\ \therefore \quad T_f & =\Delta T+T=675 \mathrm{~K} \end{aligned} $
Further at constant volume
$ \begin{aligned} \frac{p_2}{p_1} & =\frac{T_2}{T_1} \\ \therefore \quad p_2 & =\frac{T_2}{T_1} p_1 \\ & =\left(\frac{675}{300}\right)\left(1.6 \times 10^6\right) \\ & =3.6 \times 10^6 \mathrm{~N} / \mathrm{m}^2 \end{aligned} $
BETA
