A monoatomic gas (gamma = 5/3) at a temperatu | vedclass

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(B) For an adiabatic process,the relation between temperature and volume is given by $T V^{\gamma - 1} = 	ext{constant}$.Initial temperature $T_1 = 2

Vedclass · Accepted Answer

$375$

Vedclass · Answer

(B) For an adiabatic process,the relation between temperature and volume is given by $T V^{\gamma - 1} = 	ext{constant}$.Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.Given volume change: $V_2 = \frac{8}{27} V_1$,so $\frac{V_1}{V_2} = \frac{27}{8}$.Using the adiabatic relation: $T_2 = T_1 \left( \frac{V_1}{V_2} ight)^{\gamma - 1}$.Substituting the values: $T_2 = 300 \left( \frac{27}{8} ight)^{\frac{5}{3} - 1} = 300 \left( \frac{27}{8} ight)^{2/3}$.$T_2 = 300 \left( \left( \frac{3}{2} ight)^3 ight)^{2/3} = 300 \left( \frac{3}{2} ight)^2 = 300 	imes \frac{9}{4} = 675 \ K$.The change in temperature is $\Delta T = T_2 - T_1 = 675 - 300 = 375 \ K$.

$A$ monoatomic gas $(\gamma = 5/3)$ at a temperature of $27^{\circ}C$ is compressed adiabatically to $8/27$ of its initial volume. The change in temperature is . . . . . . $K$.

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