A cylinder contains 1.50 moles of an ideal monatomic gas that is initially at a temperature of 317 K. If the gas gains 2730 J of heat and performs 780 J of work, what is its final temperature? (R = 8.31 J/mol ∙ K)
A) 359 K
B) 421 K
C) 526 K
D) 687 K
E) 756 K

Question

Quizplus · Accepted Answer

Using the first law of thermodynamics, ΔU = Q - W. The change in internal energy (ΔU) is 2730 J - 780 J = 1950 J. For a monatomic ideal gas, ΔU = (3/2)nRΔT. Solving for ΔT gives ΔT = ΔU / ((3/2)nR) = 1950 / ((3/2) * 1.50 * 8.31) = 104 K. The final temperature is 317 K + 104 K = 421 K.

A Cylinder Contains 1