Initially,the pressure of 1 text{ mole} of an | vedclass

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(B) Given: $P_1 = 10^5 	ext{ Nm}^{-2}$,$V_1 = 16 	ext{ L} = 16 	imes 10^{-3} 	ext{ m}^3$,$V_2 = 2 	ext{ L} = 2 	imes 10^{-3} 	ext{ m}^3$,$C_v =

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$7.2$

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(B) Given: $P_1 = 10^5 	ext{ Nm}^{-2}$,$V_1 = 16 	ext{ L} = 16 	imes 10^{-3} 	ext{ m}^3$,$V_2 = 2 	ext{ L} = 2 	imes 10^{-3} 	ext{ m}^3$,$C_v = \frac{3R}{2}$.For an ideal gas,$C_p = C_v + R = \frac{3R}{2} + R = \frac{5R}{2}$.The adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3}$.For an adiabatic process,$P_1 V_1^\gamma = P_2 V_2^\gamma$.$P_2 = P_1 \left( \frac{V_1}{V_2} ight)^\gamma = 10^5 \left( \frac{16}{2} ight)^{5/3} = 10^5 	imes (8)^{5/3} = 10^5 	imes (2^3)^{5/3} = 10^5 	imes 2^5 = 32 	imes 10^5 	ext{ Nm}^{-2}$.Work done by the gas $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.$W = \frac{(10^5 	imes 16 	imes 10^{-3}) - (32 	imes 10^5 	imes 2 	imes 10^{-3})}{5/3 - 1} = \frac{1600 - 6400}{2/3} = \frac{-4800}{2/3} = -4800 	imes \frac{3}{2} = -7200 	ext{ J} = -7.2 	ext{ kJ}$.Since the work done by the gas is $-7.2 	ext{ kJ}$,the work done on the gas is $7.2 	ext{ kJ}$.

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