Question

An ideal gas has specific heat capacity at constant pressure \(\dfrac{11}{10}R\). If one mole of this ideal gas at \(125^\circ\text{C}\) does \(83\,\text{J}\) of work adiabatically, then the final temperature of the gas would be \((R=8.3\,\text{J K}^{-1}\text{mol}^{-1})\)

• A. \(25^\circ\text{C}\)
• B. \(50^\circ\text{C}\)
• C. \(75^\circ\text{C}\)
• D. \(100^\circ\text{C}\)

Hint:

For an adiabatic process, \[ Q=0 \] and \[ \Delta U=-W. \] Also remember that for an ideal gas, \[ C_P-C_V=R. \] These two relations are sufficient to solve most adiabatic temperature-change problems.

Answer 1

The Correct Option is A

Step 1: Determine \(C_V\) of the gas.
Given, \[ C_P=\frac{11}{10}R \] For an ideal gas, \[ C_P-C_V=R \] Therefore, \[ C_V=C_P-R \] \[ C_V=\frac{11}{10}R-R \] \[ C_V=\frac{1}{10}R \] Substituting \[ R=8.3\,\text{J K}^{-1}\text{mol}^{-1}, \] \[ C_V=\frac{8.3}{10} \] \[ C_V=0.83\,\text{J K}^{-1}\text{mol}^{-1} \]

Step 2: Apply the first law of thermodynamics for an adiabatic process.
For an adiabatic process, \[ Q=0 \] Hence, \[ \Delta U=-W \] Given, \[ W=83\,\text{J} \] Therefore, \[ \Delta U=-83\,\text{J} \]

Step 3: Express \(\Delta U\) in terms of temperature.
For one mole of an ideal gas, \[ \Delta U=nC_V(T_2-T_1) \] Since \[ n=1, \] \[ 0.83(T_2-T_1)=-83 \] \[ T_2-T_1=-100 \] \[ T_2=T_1-100 \]

Step 4: Substitute the initial temperature.
Initial temperature is \[ 125^\circ\text{C} \] or \[ T_1=398\,\text{K} \] Hence, \[ T_2=398-100 \] \[ T_2=298\,\text{K} \] Converting back to Celsius, \[ T_2=298-273 \] \[ T_2=25^\circ\text{C} \]

Step 5: Final conclusion.
Therefore, the final temperature of the gas is \[ \boxed{25^\circ\text{C}} \]