Question 848: Thermodynamics & KTG

Question

A gas at normal temperature is suddenly compressed to one-fourth of its original volume. If $$\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma=1.5$$, then the increase in its temperature is

Accepted Answer

Given that, $$V_2=\frac{V_1}{4}, \frac{C_p}{C_v}=\gamma=1.5$$
As the process is sudden, it is an adiabatic expansion,
$$\begin{aligned}
& 	herefore \quad \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \
& 	herefore \quad \mathrm{T}_2=\mathrm{T}_1\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}ight)^{\gamma-1} \
& =\mathrm{T}_1(4)^{\gamma-1} \
& =\mathrm{T}_1 	imes(4)^{0.5} \
& =2 \mathrm{~T}_1 \
& \mathrm{~T}_2-\mathrm{T}_1=\mathrm{T}_1 \
& \mathrm{~T}_2-\mathrm{T}_1=273 \mathrm{~K} \quad\left(\because \mathrm{T}_1=	ext { Normal temperature }ight) \
\end{aligned}$$

Answer

273 K

Answer

373 K

Answer

473 K

Answer

573 K

Thermodynamics & KTG

A gas at normal temperature is suddenly compressed to one-fourth of its original volume. If $\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\gamma=1.5$, then the increase in its temperature is

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