A vessel contains 0.15 text{m}^3 of a gas at | vedclass

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(B) For an adiabatic process, the adiabatic index $\gamma = c_p / c_v = 3R / 2R = 1.5$.Initial state: $P_i = 8 \ 	ext{bar} = 8 	imes 10^5 \ 	ext{Pa

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) For an adiabatic process, the adiabatic index $\gamma = c_p / c_v = 3R / 2R = 1.5$.Initial state: $P_i = 8 \ 	ext{bar} = 8 	imes 10^5 \ 	ext{Pa}$, $V_i = 0.15 \ 	ext{m}^3$.Final state: $P_f = 1 \ 	ext{bar} = 1 	imes 10^5 \ 	ext{Pa}$.Using the adiabatic relation $P_i V_i^\gamma = P_f V_f^\gamma$, we find the final volume $V_f$:$V_f = V_i (P_i / P_f)^{1/\gamma} = 0.15 	imes (8/1)^{1/1.5} = 0.15 	imes (8)^{2/3} = 0.15 	imes 4 = 0.6 \ 	ext{m}^3$.The work done in an adiabatic process is given by $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.$W = \frac{(8 	imes 10^5 	imes 0.15) - (1 	imes 10^5 	imes 0.6)}{1.5 - 1} = \frac{120000 - 60000}{0.5} = \frac{60000}{0.5} = 120000 \ 	ext{J} = 120 \ 	ext{kJ}$.

$A$ vessel contains $0.15 \ \text{m}^3$ of a gas at pressure $8 \ \text{bar}$ and temperature $140^\circ \text{C}$ with $c_p = 3R$ and $c_v = 2R$. It is expanded adiabatically until the pressure falls to $1 \ \text{bar}$. The work done during this process is . . . . . . $\text{kJ}$.

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