Question

(17%) Problem 4: A monatomic ideal gas is in a state with volume of Vo at pressure Po and temperature T . The following questions refer to the work done on the gas, W- -PA 17% Part (a) The gas undergoes an isochoric cooling from its initial state (I-Po-T0). For this process, choose what happens to the energy heat, and work from the following Grade Summary Deductions Potential 100% 0% Submissions OAU > 0, Δυ-o-w. Q < 0, and w > 0 Attempts remaining:2 (112 per attempl) detailed view Δυ < 0, Δυ-w, and Q-0 None of these. OaU < 0,AU = Q, and W = 0 Hint I give up! Feedback: 1% deduction per feedback. Hints: 0%-deduction per hint. Hints remaining ㅿ 17% Part (b) The gas undergoes an isobaric expansion from its initial state (1 -Pah For this process, choose what happens to the energy, heat, and work from the following ▲ 17% Part (c) T he gas undergoes an isothermal expansion from, its initial state (I o Po Ta). For this process, choose what happens to the energy, heat, and work from the following ▲ 17% Part (d) The gas undergoes an adiabatic expansion fromíts initial state (Vo Po-To For this process, choose what happens to the energy heat, and work from the following ǐ 17% Part (e) The gas undergoes an adiabatic expansion from its initial state (Vh Po-TO) The expansion is followed by an sobanc compression, an isochoric heating, and an isothermal expansion taking the gas back to its original state. Choose the PV diagram that shows this process. 몰 ▲ 17% Part (f) In the cycle described by part (e), choose what happens to the energy, heat, and work from the following:

Answer 1

(a) Isochoric cooling theta < 0 Volume remain constant v = constant: Delta v = 0 So w = -p dv = 0 From First law of thermodynamics Delta u = theta + w Delta u = theta As theta < 0 So Delta u < 0 (b) Isobaric expansion i.e. pressure remain constant in expansion Delta v > 0 i.e. volume increases w = -p dv = -p dv Work is done on the system So, w < 0 from thermodynamics First law Delta u = theta + w In expansion theta < 0 So Delta u < 0 (c) In isothermal expansion temperature remain constant Delta T = 0 So, Delta u = 0 From First law of thermodynamics Delta u = theta + w 0 = theta + w theta = -w in expansion theta < 0 (d) In adiabatic expansion heat: theta = 0 From First law of thermodynamics Delta u = theta