Isochoric and Isobaric (Ch.15 problem #26 Wiley+)

[trinot]

Homework Statement

Chapter 15, Problem 26

The drawing refers to 5.30 mol of a monatomic ideal gas and shows a process that has four steps, two isobaric (A to B, C to D) and two isochoric (B to C, D to A). (a) What is the work done from A to B? (b) What is the heat added or removed from B to C? (c) What is the change in internal energy from C to D? (d) What is the work done from D to A?

Homework Equations

W= nRT(Vf/Vi)

The Attempt at a Solution

W= 5.3mol * 8.31 * 800K *(Vf/Vi) this is where I am stick I don't know how to find the volume. I tried using V=nRT/P but I don't know the pressure?

Thanks in advance, the homework is due tonight.

 

[Hootenanny]

Welcome to PF trinot,

W= 5.3mol * 8.31 * 800K *(Vf/Vi) this is where I am stick I don't know how to find the volume. I tried using V=nRT/P but I don't know the pressure?

No you don't know the actual value of the pressure, but you know that it remains constant throughout the process

 

[trinot]

I am still not sure what to do, does this mean I can ignore it and just use V=nRT/P?

Thanks!

 

[alphysicist]

Hi trinot,

I believe you are using the wrong equation for the work done by the gas. For an isothermal process the work done by a gas is

<br /> W = nRT \ln\left({\frac{V_f}{V_i}\right)<br />

but these processes are not isothermal.

For an isochoric process the work done is zero; for an isobaric process the work done by the gas is

<br /> W= P (V_f - V_i)<br />

Since you know n and T for all four points on the graph you can find the values of (PV_f) and (PV_i).