Q.47. A vessel contains of a gas at pressure 8 bar and temperature 140°C with and . It is expanded

Q: What is the correct answer for this exam Physics Q47?

The correct answer is 120. Sol. Given $V_1 = 0.15 \text{ m}^3$, $P_1 = 8 \text{ bar} = 8 \times 10^5 \text{ Pa}$. $T_1 = 140^\circ C = 140 + 273 = 413 \text{ K}$. $P_2 = 1 \text{ bar} = 1 \times 10^5 \text{ Pa}$. $C_p = 3R$, $C_v = 2R$. The adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{3R}{2R} = 1.5$. For an adiabatic process, $P_1 V_1^\gamma = P_2 V_2^\gamma$. So $V_2 = V_1 \left(\frac{P_1}{P_2}\right)^{1/\gamma} = 0.1

Q: Q.47. A vessel contains of a gas at pressure 8 bar and temperature 140°C with and . It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is _______ kJ. (R is…

The correct answer is 120. Sol. Given $V_1 = 0.15 \text{ m}^3$, $P_1 = 8 \text{ bar} = 8 \times 10^5 \text{ Pa}$. $T_1 = 140^\circ C = 140 + 273 = 413 \text{ K}$. $P_2 = 1 \text{ bar} = 1 \times 10^5 \text{ Pa}$. $C_p = 3R$, $C_v = 2R$. The adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{3R}{2R} = 1.5$. For an adiabatic process, $P_1 V_1^\gamma = P_2 V_2^\gamma$. So $V_2 = V_1 \left(\frac{P_1}{P_2}\right)^{1/\gamma} = 0.1

Question

Q.47. A vessel contains  of a gas at pressure 8 bar and temperature 140°C with  and . It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is _______ kJ. (R is gas constant)

FirstInTest · Accepted Answer

Sol. Given $V_1 = 0.15 	ext{ m}^3$, $P_1 = 8 	ext{ bar} = 8 	imes 10^5 	ext{ Pa}$. $T_1 = 140^\circ C = 140 + 273 = 413 	ext{ K}$. $P_2 = 1 	ext{ bar} = 1 	imes 10^5 	ext{ Pa}$. $C_p = 3R$, $C_v = 2R$. The adiabatic index $\gamma = \frac{C_p}{C_v} = \frac{3R}{2R} = 1.5$. For an adiabatic process, $P_1 V_1^\gamma = P_2 V_2^\gamma$. So $V_2 = V_1 \left(\frac{P_1}{P_2}ight)^{1/\gamma} = 0.15 \left(\frac{8 	imes 10^5}{1 	imes 10^5}ight)^{1/1.5} = 0.15 (8)^{2/3} = 0.15 	imes (2^3)^{2/3} = 0.15 	imes 2^2 = 0.15 	imes 4 = 0.6 	ext{ m}^3$. The work done in an adiabatic process is $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$. $W = \frac{(8 	imes 10^5 	imes 0.15) - (1 	imes 10^5 	imes 0.6)}{1.5 - 1} = \frac{(1.2 	imes 10^5) - (0.6 	imes 10^5)}{0.5} = \frac{0.6 	imes 10^5}{0.5} = 1.2 	imes 10^5 	ext{ J} = 120 	ext{ kJ}$.

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