## A monatomic ideal gas has pressure p1 and temperature T1. It is contained in a cylinder of volume V1 with a movable piston, so that it can d

Question

A monatomic ideal gas has pressure p1 and temperature T1. It is contained in a cylinder of volume V1 with a movable piston, so that it can do work on the outside world.Consider the following three-step transformation of the gas:The gas is heated at constant volume until the pressure reaches Ap1 (where A>1).The gas is then expanded at constant temperature until the pressure returns to p1.The gas is then cooled at constant pressure until the volume has returned to V1.It may be helpful to sketch this process on the pVplane.How much heat Q1 is added to the gas during step 1 of the process?Express the heat added in terms of p1, V1, and A.How much work W2 is done by the gas during step 2?Express the work done in terms of p1, V1, and A.How much work W3 is done by the gas during step 3?If you’ve drawn a graph of the process, you won’t need to calculate an integral to answer this question.Express the work done in terms of p1, V1, and A.

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1 month 2021-08-13T21:07:24+00:00 1 Answers 0 views 0

A) Q1 = (3/2)P1V1[A – 1]

B) W2 = P1V1(In A)

C) W3 = P1V1(1 – A)

Explanation:

A) From first law of thermodynamics and applying to the question, we have;

ΔU = Q – W

Where,

ΔU = change in internal energy

Q = the heat absorbed

W = the work done

Now, because the first process occurs at constant volume, the work done is zero:

Thus,

ΔU = Q – 0

ΔU = Q

The change in internal energy is given by;

ΔU = nCvΔt

where;

n = the number of moles of the gas

R = the gas constant,

Cv = the specific heat at constant volume

Δt = The change in temperature i.e T2 – T1.

Now, using the ideal gas law, let us find an expression for n and Δt

P1V1 = nRT1

n = P1V1/RT1

T1 = P1V1/nR

Now, the specific heat at constant volume is Cv = (3/2)R

Now, from the question, since it’s pressure has reached AP1, we can calculate the temperature T2 by using the ideal gas law at the new conditions of the gas as;

AP1V1 = nRT2

T2 = AP1 V1/ nR

Now, we are to express the heat added in terms of p1, V1, and A

Q = ΔU = nCv(T2 – T1)

From earlier, we saw that,

T1 = P1V1/nR

Putting equation of T2 and T1 into the energy equation to get;

Q = nCv((AP1 V1/ nR) – P1V1/nR)

Q = Cv • P1V1/R (A – 1)

Now, from earlier, we saw that Cv = (3/2)R. Thus,

Q = (3/2)R • P1V1/R (A – 1)

Q = (3/2)P1V1[A – 1]

B) Here again, we are to express work done in step 2 in terms of p1, V1, and A.

This process is an isothermal process because temperature is constant and so work done is given as; W = nRT In(V2/V1)

T = T1 because temperature is constant

From earlier, we saw that;

n = P1V1/RT1 and

But in this process, it’s

n = P1V1/RT1 and thus,

V2 = nRT2/P1

We also saw that T2 = AP1 V1/ nR

V1 = nRT2/AP1

Plugging in the relevant values into, W = nRT In(V2/V1), we obtain;

W = (P1V1/RT1) • RT1 • In((nRT2/P1)/(nRT2/AP1)

W = P1V1(In A)

C) In step 3,we have and isobaric process because the pressure is constant.

Work done in this case is given by ;

W = P(V1 – V2)

Because V2 in now the final volume while V1 is now the the initial volume

Now, P is P1 because it’s an isobaric process.

From earlier, we saw that,

V1 = nRT2/AP1 and V2 = nRT2/P1

And that T2 = AP1 V1/ nR

Thus,

V1 = V1 and V2 = AV1

Thus, W = P1(V1 – AV1) = P1V1(1 – A)