Question

The temperature of \(2\) moles of a monoatomic gas changes from \(25^{\circ}C\) to \(35^{\circ}C\) in an adiabatic process. Find the work done.

• A. \(-249.4\,J\)
• B. \(+249.4\,J\)
• C. \(-498.8\,J\)
• D. \(+498.8\,J\)

Hint:

For an adiabatic process, \[ \boxed{Q=0} \] and \[ \boxed{W=-\Delta U.} \] For a monoatomic ideal gas, \[ \boxed{C_V=\frac{3R}{2}} \] Hence, \[ \boxed{W=-\frac32nR\Delta T.} \] Always calculate the temperature difference in Kelvin. Since temperature intervals are the same in Celsius and Kelvin, \[ \boxed{\Delta T(^\circ C)=\Delta T(K).} \]

Answer 1

The Correct Option is A

At constant volume, heat absorbed is $Q = nC_v\Delta T$ where $C_v = \frac{3}{2}R$ for a monoatomic ideal gas. For $n = 2$ mol and $\Delta T = 10$ K: $Q = 2 \times \frac{3}{2}R \times 10 = 30R \approx 249$ J.