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

  • Thread starter trinot
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  • #1
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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 im 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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

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Answers and Replies

  • #2
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
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Welcome to PF trinot,
W= 5.3mol * 8.31 * 800K *(Vf/Vi) this is where im 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 :wink:
 
  • #3
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I am still not sure what to do, does this mean I can ignore it and just use V=nRT/P?

Thanks!
 
  • #4
alphysicist
Homework Helper
2,238
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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

[tex]
W = nRT \ln\left({\frac{V_f}{V_i}\right)
[/tex]

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

[tex]
W= P (V_f - V_i)
[/tex]

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

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