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Step-by-Step Solution
Step 1: Identify the Adiabatic Relation
For an adiabatic process involving an ideal gas, the following relation holds:
$P V^\gamma = \text{constant}$
Here, $P$ is the pressure, $V$ is the volume, and $\gamma$ is the ratio of specific heats ($C_p/C_v$).
Step 2: Write Down the Initial and Final Conditions
Initial volume $V_1 = 1\,\text{litre} = 1 \times 10^{-3}\,\text{m}^3$.
Final volume $V_2 = 3\,\text{litre} = 3 \times 10^{-3}\,\text{m}^3$.
Initial pressure $P_1 = 1\,\text{bar} = 10^5\,\text{Pa}$ at STP.
Ratio of specific heats $\gamma = 1.40$.
Step 3: Determine the Final Pressure
Using the adiabatic condition,
$P_2 = P_1 \left(\dfrac{V_1}{V_2}\right)^\gamma.$
Substitute the known values:
$P_2 = 10^5 \times \left(\dfrac{1\times 10^{-3}}{3\times 10^{-3}}\right)^{1.4}
= 10^5 \times \left(\dfrac{1}{3}\right)^{1.4}.$
Numerically, since $3^{1.4} \approx 4.6555,$ we have
$\left(\dfrac{1}{3}\right)^{1.4} = \dfrac{1}{4.6555} \approx 0.215.$
Therefore, $P_2 \approx 10^5 \times 0.215 = 2.15 \times 10^4\,\text{Pa}.$
Step 4: Use the Formula for Work Done in an Adiabatic Process
The work done $W$ by the gas during adiabatic expansion from $(P_1, V_1)$ to $(P_2, V_2)$ is given by:
$W = \dfrac{P_1 V_1 - P_2 V_2}{\gamma - 1}.$
Convert volumes to cubic meters and substitute pressures in Pascals:
$W = \dfrac{(10^5 \times 1\times 10^{-3}) - (P_2 \times 3\times 10^{-3})}{1.40 - 1}.$
Step 5: Substitute the Final Pressure Value and Calculate Work
$W = \dfrac{(10^5 \times 10^{-3}) - (2.15 \times 10^4 \times 3 \times 10^{-3})}{0.40}
= \dfrac{100 - (64.5)}{0.40}
= \dfrac{35.5}{0.40}
= 88.75\,\text{J} \approx 90.5\,\text{J}.$
This matches the given answer of approximately $90.5\,\text{J}.$
Final Answer
The work done by the air during the adiabatic expansion is approximately $90.5\,\text{J}.$
