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Adiabatic process
Appears in
Concepts tested here
- Adiabatic compression
- Pressure-temperature relation
- P–T relation in adiabatic process
All Questions
2013 NEET 1 question
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During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its temperature. The ratio of $\frac{C_p}{C_v}$ for the gas is:$P \propto T^3$ and $PV = nRT$ give $PV^{3/2}$ = constant.
Comparing with $PV^\gamma$ = constant: $\gamma = \frac{C_p}{C_v} = \frac{3}{2}$
2011 AIPMT-MAINS 1 question
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A mass of diatomic gas ($\gamma = 1.4$) at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from $27^\circ C$ to $927^\circ C$. The pressure of the gas in the final state is$T_1 = 27 + 273 = 300$ K; $T_2 = 927 + 273 = 1200$ K; $P_1 = 2$ atm.
For an adiabatic process $P^{1-\gamma}T^\gamma$ = constant, so $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma - 1}}$
$\frac{\gamma}{\gamma - 1} = \frac{1.4}{0.4} = \frac{7}{2}$
$P_2 = 2\times\left(\frac{1200}{300}\right)^{7/2} = 2\times(4)^{7/2}$
$(4)^{7/2} = 2^7 = 128$
$P_2 = 2\times128 = 256$ atm
2010 AIPMT-MAINS 1 question
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A monoatomic gas at pressure $P_1$ and volume $V_1$ is compressed adiabatically to $\frac{1}{8}$th its original volume. What is the final pressure of the gas?For an adiabatic process: $PV^\gamma$ = constant; for a monoatomic gas $\gamma = \frac{5}{3}$
$P_1V_1^{5/3} = P_2\left(\frac{V_1}{8}\right)^{5/3}$
$P_2 = P_1(8)^{5/3}$
$(8)^{5/3} = (2^3)^{5/3} = 2^5 = 32$
$P_2 = 32P_1$