Question 92: Thermodynamics & KTG

Question

Initial pressure and volume of a monoatomic ideal gas are $P$ and $V$. The change in internal energy of this gas in adiabatic expansion to volume $V_{	ext {final }}=27 \mathrm{~V}$ is $\_\_\_\_$ J.

Accepted Answer

In an adiabatic process, no heat is exchanged with the surroundings ( $\mathrm{Q}=0$ ).
For an ideal gas undergoing an adiabatic change, the relationship between pressure and volume is :
$$ \mathrm{PV}^\gamma=	ext { constant } $$
Where $\gamma$ is the ratio of specific heats $\left(\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}ight)$.
For a monoatomic gas, $\gamma=53$.
The initial pressure of gas is $\mathrm{P}_1=\mathrm{P}$.
The initial volume of gas is $\mathrm{V}_1=\mathrm{V}$
The final volume of gas is $V_2=27 \mathrm{~V}$
Using the adiabatic relation $\mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma$ :
$$ \mathrm{P} \cdot \mathrm{~V}^{5 / 3}=\mathrm{P}_{	ext {final }} \cdot(27 \mathrm{~V})^{5 / 3} $$
$$\Rightarrow $$ $\mathrm{P}_{	ext {final }}=\mathrm{P}\left(\frac{\mathrm{V}}{27 \mathrm{~V}}ight)^{5 / 3}=\mathrm{P}\left(\frac{1}{27}ight)^{5 / 3}$
$$ \mathrm{P}_{	ext {final }}=\mathrm{P}\left(\frac{1}{3^3}ight)^{5 / 3}=\mathrm{P}\left(\frac{1}{3}ight)^5=\frac{\mathrm{P}}{243} $$
From the $1^{	ext {st }}$ law of thermodynamics $(Q=\Delta U+W)$, since $Q=0$, we have:
$$ \Delta U=-W $$
The work done ( W ) in an adiabatic process is:
$$ \mathrm{W}=\frac{\mathrm{P}_1 \mathrm{~V}_1-\mathrm{P}_2 \mathrm{~V}_2}{\gamma-1} $$
Therefore:
$$ \Delta U=\frac{P_2 V_2-P_1 V_1}{\gamma-1} $$
Substitute the given values, $\mathrm{P}_1 \mathrm{~V}_1=\mathrm{PV}, \mathrm{P}_2 \mathrm{~V}_2=\left(\frac{\mathrm{P}}{243}ight) 	imes(27 \mathrm{~V})=\frac{\mathrm{PV}}{9}$
$\Delta U=\frac{\frac{P V}{9}-P V}{\frac{5}{3}-1}$
$$\Rightarrow $$ $\Delta U=\frac{-\frac{8}{9} P V}{2 / 3}$
$$\Rightarrow $$ $\Delta \mathrm{U}=-\frac{8}{9} \mathrm{PV} 	imes \frac{3}{2}$
$$\Rightarrow $$ $$ \Delta \mathrm{U}=-\frac{4}{3} \mathrm{PV} $$
Therefore, the change in internal energy is $-\frac{4}{3} \mathrm{PV}$.
Hence, the correct option is (C).

Answer

$-2 P V(3 \sqrt{3}-1)$

Answer

$\frac{4}{3} P V$

Answer

$-\frac{4}{3} P V$

Answer

$$ \frac{3}{4} P V $$

Thermodynamics & KTG

Initial pressure and volume of a monoatomic ideal gas are P and V. The change in internal energy of this gas in adiabatic expansion to volume V_{\text {final }}=27 \mathrm{~V} is \_\_\_\_ J.

Choose the correct answer: