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 from given C_P.
C_V = C_P – R = (11/10)R – R = R/10 = 0.83 J/(mol·K).

Step 2: Apply first law for adiabatic process (Q = 0).
ΔU = –W = –83 J. Also ΔU = nC_VΔT = 0.83(T₂ – T₁).

Step 3: Solve for temperature change.
0.83(T₂ – T₁) = –83 → T₂ – T₁ = –100 K. T₁ = 125°C = 398 K → T₂ = 298 K = 25°C.

Step 4: Final Answer:
25°C.