Question

In an adiabatic expansion, the temperature of one mole of an ideal monoatomic gas \((\gamma=\frac{5}{3})\) decreases from \(60\,\text{K}\) to \(50\,\text{K}\). The work done by the gas in the process is: (Take the universal gas constant as \(R=8.3\,\text{J mol}^{-1}\text{K}^{-1}\))

• A. \(166\,\text{J}\)
• B. \(41.5\,\text{J}\)
• C. \(83\,\text{J}\)
• D. \(124.5\,\text{J}\)

Hint:

For adiabatic processes, heat exchange is zero. Work done equals decrease in internal energy. For monoatomic gases, \(C_V=\frac{3R}{2}\). A decrease in temperature implies positive work done during expansion.

Answer 1

The Correct Option is D

Step 1: State the process.
One mole of a monoatomic ideal gas ($\gamma = 5/3$) expands adiabatically as its temperature drops from $60\,\text{K}$ to $50\,\text{K}$, with $R = 8.3\,\text{J mol}^{-1}\text{K}^{-1}$.
Step 2: Adiabatic condition.
In an adiabatic process no heat flows, so $Q = 0$.
Step 3: First law.
The first law $Q = \Delta U + W$ then gives $W = -\Delta U$.
Step 4: Molar heat capacity.
For a monoatomic gas $C_V = \tfrac{3}{2}R$, so $\Delta U = n C_V \Delta T = \tfrac{3}{2}R\,(T_2 - T_1)$.
Step 5: Evaluate the internal energy change.
\[ \Delta U = \tfrac{3}{2}(8.3)(50 - 60) = 1.5 \times 8.3 \times (-10) = -124.5\,\text{J} \]
Step 6: Work done by the gas.
Since $W = -\Delta U$, the gas does positive work as it cools.
\[ W = -(-124.5) = 124.5\,\text{J} \]
\[ \boxed{124.5\,\text{J}} \]