NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
A gas is allowed to expand at constant temperature from a volume of 1.0 L to 10.1 L against an external pressure of 0.50 atm. If the gas absorbs 250 J of heat from the surroundings, what are the values of q, w and Δ E ? (Given 1 L atm = 101.0 J)
Options
- Aq w Δ E 250 J - 460 J - 210 J
- Bq w Δ E -250 J - 460 J - 710 J
- Cq w Δ E 250 J 460 J 710 J
- Dq w Δ E -250 J 460 J 210 J
Correct answer
A. q w Δ E 250 J - 460 J - 210 J
Step-by-step solution
Since heat is absorbed, q = + 250 J Work done = - p V 2 - V 1 = - 50 atm × 10.1 L - 1.0 L = - 0.50 atm × 9.1 L = - 0.50 × 9.1 L atm × 101.0 J 1 L atm =- 460.0 J Also Δ E = q + w = 250 J - 460 J = - 210 J