The temperature of n moles of an ideal monatomic gas is increased by $\Delta$ T at constant volume.The energy Q absorbed as heat, change $\Delta$E_int in internal energy, and work W done by the environment are given by: A)Q = (5/2)nR$\Delta$T, $\Delta$E_int = 0, W = 0 B)Q = (3/2)nR$\Delta$T, $\Delta$E_int = (3/2)nR$\Delta$T, W = 0 C)Q = (3/2)nR$\Delta$T, $\Delta$E_int = (1/2)nR$\Delta$T, W = -nR$\Delta$T D)Q = (5/2)nR$\Delta$T, $\Delta$E_int = (3/2)nR$\Delta$T, W = -nR$\Delta$T E)Q = (3/2)nR$\Delta$T, $\Delta$E_int = 0, W = -(3/2)nR$\Delta$T

Question

Quizplus · Accepted Answer

For an ideal monatomic gas at constant volume, the heat absorbed (Q) increases the internal energy (ΔU) with no work done (W=0), and the molar specific heat at constant volume (Cv) is (3/2)R, making Q and ΔU equal to (3/2)nRΔT.

The Temperature of N Moles of an Ideal Monatomic Gas Δ\DeltaΔ

The Temperature of N Moles of an Ideal Monatomic Gas $\Delta$