A monatomic gas at a pressure of 100 text{ kP | vedclass

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(A) For an adiabatic process, the work done $W$ is given by the formula: $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.Since $P_i V_i^\gamma = P_f V_f^\g

Vedclass · Accepted Answer

$1600$

Vedclass · Answer

(A) For an adiabatic process, the work done $W$ is given by the formula: $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.Since $P_i V_i^\gamma = P_f V_f^\gamma$, we have $P_f = P_i (V_i / V_f)^\gamma$.Given $V_f = 8 V_i$ and for a monatomic gas $\gamma = 5/3$, we get $P_f = P_i (1/8)^{5/3} = P_i / 32$.Substituting these into the work formula: $W = \frac{P_i V_i - (P_i / 32)(8 V_i)}{5/3 - 1} = \frac{P_i V_i - P_i V_i / 4}{2/3} = \frac{(3/4) P_i V_i}{2/3} = \frac{9}{8} P_i V_i$.Given $W = 180 	ext{ J}$ and $P_i = 100 	imes 10^3 	ext{ Pa}$, we have $180 = \frac{9}{8} 	imes 10^5 	imes V_i$.$V_i = \frac{180 	imes 8}{9 	imes 10^5} = \frac{160}{10^5} = 1.6 	imes 10^{-3} 	ext{ m}^3$.Converting to $	ext{cm}^3$: $1.6 	imes 10^{-3} 	imes 10^6 	ext{ cm}^3 = 1600 	ext{ cm}^3$.

$A$ monatomic gas at a pressure of $100 \text{ kPa}$ expands adiabatically such that its final volume becomes $8$ times its initial volume. If the work done during the process is $180 \text{ J}$, then the initial volume of the gas is (in $\text{ cm}^3$)

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