Practice free →
Home › NEET UG › Physics › Thermodynamics › One mole of ideal gas with γ = 1.4 cools by 20 K…

One mole of ideal gas with γ = 1.4 cools by 20 K in adiabatic expansion. With R = 8.3 J mol^-1 K^-1, the work done is

A249 J
B415 J
C664 J
D166 J
Answer & Solution
Correct answer: B. 415 J
W = µR(T1 - T2)/(γ - 1) = 1 x 8.3 x 20 / 0.4 = 415 J.
Solve this in the app — NEET UG practice & 24k+ MCQs →
Related questions