Question 873: Thermodynamics & KTG

Question

A sample of gas at temperature $$T$$ is adiabatically expanded to double its volume. The work done by the gas in the process is $$\left(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{3}{2}\right) \quad(\mathrm{R}=$$ gas constant $$)$$

Accepted Answer

Using the formula for adiabatic expansion,
$$\begin{aligned}
& \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \
& \mathrm{~T}_1 \mathrm{~V}_1{ }^{(1 / 2)}=\mathrm{T}_2\left(2 \mathrm{~V}_1{ }^{1 / 2}ight) \quad ..... (	ext{Given}: \gamma=\frac{3}{2})\
	herefore \quad & \mathrm{T}_1=\mathrm{T}_2(\sqrt{2}) \
	herefore \quad & \mathrm{T}_2=\frac{\mathrm{T}}{\sqrt{2}}
\end{aligned}$$
Work done
$$\begin{aligned}
\mathrm{W}_{\mathrm{adi}}= & \frac{\mathrm{R}\left(\mathrm{T}-\mathrm{T}_2ight)}{\gamma-1}=\frac{\mathrm{R}\left(\mathrm{T}-\frac{\mathrm{T}}{\sqrt{2}}ight)}{\frac{1}{2}} \
& =\frac{\mathrm{R}(\sqrt{2} \mathrm{~T}-\mathrm{T})}{\sqrt{2}} 	imes 2=\mathrm{RT}(\sqrt{2}-1) \sqrt{2} \
	herefore \quad \mathrm{W}_{	ext {adi }} & =\mathrm{RT}(2-\sqrt{2})
\end{aligned}$$

Answer

$$\operatorname{TR}(\sqrt{2}-2)$$

Answer

$$\frac{\mathrm{T}}{\mathrm{R}}(\sqrt{2}-2)$$

Answer

$$\frac{\mathrm{R}}{\mathrm{T}}(2-\sqrt{2})$$

Answer

$$\mathrm{RT}(2-\sqrt{2})$$

Thermodynamics & KTG

A sample of gas at temperature $T is adiabatically expanded to double its volume. The work done by the gas in the process is \left(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{3}{2}\right) \quad(\mathrm{R}= gas constant )

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