One mole of an ideal gas is confined to a container with a movable piston. The q

Question

One mole of an ideal gas is confined to a container with a movable piston. The questions below refer to the processes shown on the PV diagram at right. Process is a change from state Xto state Y at constant pressure. Process II is a change from state Wto state Z at a different constant pressure. 2. Rank the temperatures of states w, X, Y, and Z. If any temperatures are equal, state that explicitly. Explain. a. b. X| Process | |y In the two processes, does the piston move inward, move outward, or not move? Explain. Based on your answer to part b, state whether the following quantities are positive, negative, or zero. Explain your reasoning by referring to a force and a displacement c. Process II i. the work done on the gas during Process I (W ii. the work done on the gas during Process II (Wu) d. In Process I, is the heat transfer ,the gas positive, negative, or zero? Explain.

Explanation / Answer

2. given one mole of gas, n = 1

a. For n = 1, PV = RT

now, form plot

Px = 5, Vx = 1

Py = 5, Vy = 5

Pw = 1, Vw = 1

Pz = 1, Vz = 5

hence using T = PV/R and let R = 1 (as this is just a comparision )

Tx = 5

Ty = 25

Tw = 1

Tz = 5

hence Ty > Tx = Tz > Tw

b. in process 1 and 2, VOlume increases, so the piston moves outward

c. Process 1, work done on gas W1 < 0 as work is done by the gas and not on the gas

Process 2, work done on the gas W2 < 0 ( same reaqson as previous part)

d. Process 1 : Work done W = area under curve = 4*5 = 20

heat supplied = nCp(dT) = Cp(Ty - Tx) = Cp(20)

so heat transfer to the gas is +ve as heat supplied > 0

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