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A sample of monoatomic gas is taken at initial pressure of $75	ext{ kPa}$. The volume of the gas is then compressed from $1200	ext{ cm}^3$ to $150	ext{ cm}^3$ adiabatically. In this process, the value of work done on the gas will be: A). $79	ext{ J}$ B). $405	ext{ J}$ C). $4050	ext{ J}$ D). $9590	ext{ J}$

Edvaya · Accepted Answer

The correct answer is $405	ext{ J}$. **Solution:**
For a monoatomic gas, $\gamma = \frac{5}{3}$.
Initial state: $P_1 = 75	ext{ kPa} = 75 	imes 10^3	ext{ Pa}$, $V_1 = 1200	ext{ cm}^3$.
Final state: $V_2 = 150	ext{ cm}^3$.
Using the adiabatic relation $P_1 V_1^\gamma = P_2 V_2^\gamma$:
$P_2 = P_1 \left(\frac{V_1}{V_2}ight)^\gamma = 75 	imes 10^3 \left(\frac{1200}{150}ight)^{\frac{5}{3}} = 75 	imes 10^3 (8)^{\frac{5}{3}} = 75 	imes 10^3 	imes 32 = 2400 	imes 10^3	ext{ Pa}$.
The work done *by* the gas is $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$.
$W = \frac{(75 	imes 10^3 	imes 1200 	imes 10^{-6}) - (2400 	imes 10^3 	imes 150 	imes 10^{-6})}{\frac{5}{3} - 1} = \frac{90 - 360}{\frac{2}{3}} = \frac{-270}{\frac{2}{3}} = -405	ext{ J}$.
Since the work done *by* the gas is $-405	ext{ J}$, the work done *on* the gas is $+405	ext{ J}$.

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