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

Concept:

• In an adiabatic process, \[ Q=0 \]

• From the first law of thermodynamics, \[ \Delta U=-W \]

• For a monoatomic ideal gas, \[ C_V=\frac{3R}{2} \]

• Internal energy change is \[ \Delta U=nC_V(T_2-T_1) \]

Step 1: Write the expression for change in internal energy
\[ \Delta U = nC_V(T_2-T_1) \] For one mole, \[ n=1 \] and \[ C_V=\frac{3R}{2} \] Therefore, \[ \Delta U = \frac{3R}{2}(T_2-T_1) \]

Step 2: Substitute the given values
\[ \Delta U = \frac{3}{2}(8.3)(50-60) \] \[ \Delta U = 1.5\times8.3\times(-10) \] \[ \Delta U = -124.5\,\text{J} \]

Step 3: Apply the first law of thermodynamics
For an adiabatic process, \[ Q=0 \] Hence, \[ \Delta U=-W \] Therefore, \[ W=-\Delta U \] \[ W=124.5\,\text{J} \]

Step 4: Write the final answer
\[ \boxed{W=124.5\,\text{J}} \] Hence the correct option is \[ \boxed{\text{Option (D)}} \]