The volume of a gas is reduced adiabatically | vedclass

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(B) For an adiabatic process,the relationship between temperature and volume is given by $T V^{\gamma - 1} = 	ext{constant}$.This implies $\frac{T_2}

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$300 	imes 4^{0.4} 	ext{ K}$

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(B) For an adiabatic process,the relationship between temperature and volume is given by $T V^{\gamma - 1} = 	ext{constant}$.This implies $\frac{T_2}{T_1} = \left( \frac{V_1}{V_2} ight)^{\gamma - 1}$.Given: Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 	ext{ K}$.Initial volume $V_1 = V$.Final volume $V_2 = \frac{V}{4}$.Adiabatic index $\gamma = 1.4$.Substituting these values into the formula:$T_2 = T_1 	imes \left( \frac{V_1}{V_2} ight)^{\gamma - 1}$$T_2 = 300 	imes \left( \frac{V}{V/4} ight)^{1.4 - 1}$$T_2 = 300 	imes (4)^{0.4} 	ext{ K}$.

The volume of a gas is reduced adiabatically to $\frac{1}{4}$ of its initial volume at $27^{\circ}C$. If the value of $\gamma = 1.4$,then the new temperature will be:

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