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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
A
256 atm
B
8 atm
C
28 atm
D
68.7 atm
Detailed Solution
$T_1 = 27 + 273 = 300$ 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
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