Question 732: Thermodynamics & KTG

Question

In ideal gas of $27^{\circ} \mathrm{C}$ is compressed adiabatically to $(8 / 27)$ of its original volume. If $\gamma=\frac{5}{3}$, the rise in temperature of a gas is

Accepted Answer

For an adiabatic process, the relationship is given by:
$ \mathrm{TV}^{\gamma-1} = 	ext{constant} $
From this relationship, we can derive the expression for the temperature ratio when a volume change occurs:
$ \frac{T_2}{T_1} = \left( \frac{V_1}{V_2} ight)^{\gamma - 1} $
Given that the initial volume $V_1$ is reduced to $V_2 = \frac{8}{27} V_1$, and $\gamma = \frac{5}{3}$, the expression becomes:
$ \frac{T_2}{T_1} = \left( \frac{27}{8} ight)^{\frac{5}{3} - 1} = \left( \frac{27}{8} ight)^{\frac{2}{3}} = \frac{9}{4} $
Now, substituting the initial temperature $T_1$ (convert the Celsius temperature to Kelvin):
$ T_1 = 27 + 273 = 300 \, 	ext{K} $
Calculate $T_2$:
$ T_2 = \frac{9}{4} 	imes T_1 = \frac{9}{4} 	imes 300 = 675 \, 	ext{K} $
The rise in temperature is:
$ T_2 - T_1 = 675 - 300 = 375 \, 	ext{K} $

Answer

300 K

Answer

375 K

Answer

400 K

Answer

450 K

Thermodynamics & KTG

In ideal gas of 27^{\circ} \mathrm{C} is compressed adiabatically to (8 / 27) of its original volume. If \gamma=\frac{5}{3}, the rise in temperature of a gas is

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