Skip to main content
Back to THERMODYNAMICS

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

1 Hard
1 of 1
Application +4 | -1
A sample of argon gas at 1 atm1 \mathrm{~atm} pressure and 27∘C27^{\circ} \mathrm{C} expands reversibly and adiabatically from 1.25 dm31.25 \mathrm{~dm}^3 to 2.50 dm3.2.50 \mathrm{~dm}^3. Calculate the enthalpy change in this process. Cv,m\mathrm{C}_{\mathrm{v}, \mathrm{m}} for argon is 12.48 JK−1 mol−112.48 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}
View answer and explanation
Correct option: A
−116 J-116 \text{ J}
Option A (-116 J) is correct because it is the enthalpy change calculated using the adiabatic process equation and the given heat capacity.

More Questions from THERMODYNAMICS

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.