Question 167: Thermodynamics & KTG

Question

10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $P_1$ to $P_2$ is $\alpha$ Joule ( $P_1=21.7 \mathrm{~Pa}$ and $\left.P_2=30 \mathrm{~Pa}, \mathrm{C}_v=21 \mathrm{~J} / \mathrm{K} . \mathrm{mol}, R=8.3 \mathrm{~J} / \mathrm{mol} . \mathrm{K}\right)$. The value of $\alpha$ is $\_\_\_\_$ .

Accepted Answer

The path (AB) from state $P_1$ to state $P_2$ is a vertical line which represents isochoric process where volume remains constant.
Initial Pressure: $\mathrm{P}_1=21.7 \mathrm{~Pa}$, Initial volume: $\mathrm{V}_1=1 \mathrm{~m}^3$.
Final Pressure: $P_2=30 \mathrm{~Pa}$, Final volume: $V_2=1 \mathrm{~m}^3$
For any isochoric process, since there is no change in volume ( $\mathrm{dV}=0$ ), the work done by the gas is zero :
$$ \mathrm{W}_{\mathrm{AB}}=\int \mathrm{P} \mathrm{dV}=0 $$
Applying the $1^{	ext {st }}$ law of thermodynamics
The heat involved ( $\mathrm{Q}=\alpha$ ) in a constant volume process is equal to the change in internal energy ( $\Delta \mathrm{U}$ ) :
$$ Q=\Delta U+W $$
$$\Rightarrow $$ $$ \mathrm{Q}=\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{~T}=\alpha \ldots 	ext { (i) } $$
From the ideal gas law ( $\mathrm{PV}=\mathrm{nRT}$ ), we know that $\mathrm{nT}=\frac{\mathrm{PV}}{\mathrm{R}}$. Therefore, the change
$$ \mathrm{n} \Delta \mathrm{~T}=\frac{\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{R}} \ldots (ii)$$
From both the equations,
$$ \alpha=\mathrm{C}_{\mathrm{v}}\left(\frac{\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{R}}ight) $$
Substituting the given values :
$$ V_1=V_2=1 \mathrm{~m}^3 ; \mathrm{P}_2-\mathrm{P}_1=30-21.7=8.3 \mathrm{~Pa} ; \mathrm{C}_{\mathrm{v}}=21 \mathrm{~J} / \mathrm{Kmol} ; \mathrm{R}=8.3 \mathrm{~J} / \mathrm{molK} $$
$\alpha=21 	imes\left(\frac{(30 	imes 1)-(21.7 	imes 1)}{8.3}ight)$
$$\Rightarrow $$ $\alpha=21 	imes\left(\frac{8.3}{8.3}ight)$
Therefore, the heat involved in the process from $P_1$ to $P_2$ is 21 Joules.
Hence, the correct option is (A).

Answer

21

Answer

28

Answer

24

Answer

15

Thermodynamics & KTG

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