The temperature of n moles of an ideal monatomic gas is increased by $\Delta$T at constant pressure.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 = -nR$\Delta$T B)Q = (3/2)nR$\Delta$T, $\Delta$E_int = (5/2)nR$\Delta$T, W = -(3/2)nR$\Delta$T C)Q = (5/2)nR$\Delta$T, $\Delta$E_int = (5/2)nR$\Delta$T, W = 0 D)Q = (3/2)nR$\Delta$T, $\Delta$E_int = 0, W = -nR$\Delta$T E)Q = (5/2)nR$\Delta$T, $\Delta$E_int = (3/2)nR$\Delta$T, W = -nR$\Delta$T

Question

Quizplus · Accepted Answer

For an ideal monatomic gas, the molar specific heat at constant pressure (Cp) is (5/2)R, and the change in internal energy (∆U) for an ideal gas is given by (3/2)nR∆T. The work done by the gas at constant pressure is W = -P∆V, which can be related to the temperature change by W = -nR∆T. Therefore, the heat absorbed Q is the sum of the change in internal energy and the work done on the gas, leading to Q = (5/2)nR∆T, with ∆U = (3/2)nR∆T and W = -nR∆T.

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

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