Work done by monoatomic ideal gas

[whatisreality]

Homework Statement

The volume of 30.0 moles of a monoatomic ideal gas is reduced at a uniform rate from 0.616m³ to 0.308m³ in 2h. Its temperature is increased at a uniform rate from 27.0◦C to 450◦C. The gas passes through thermodynamic equilibrium states throughout.
(a) Write down explicitly how the temperature and the volume of the gas depend on time (in hours). Then find the dependence of T on V explicitly.
(b) Calculate the cumulative work done by the gas.
(c) Calculate the cumulative energy absorbed by the gas as heat.

Homework Equations

The Attempt at a Solution

For part a), pretty sure dV/dt = 0.154 and dT/dt = 211.5. If that's the case then making dt the subject and equation both gives
0.154 dT = 211.5 dV
b) I can't add together the work done in the volume change, then add work to change temperature, because of the path dependence of work. This is definitely an integral! Which I am really struggling to form...

 

[TSny]

For part a), pretty sure dV/dt = 0.154 and dT/dt = 211.5. If that's the case then making dt the subject and equation both gives
0.154 dT = 211.5 dV

For part (a) your expressions look OK except for a couple of sign errors. Note that the volume is decreasing while T and t increase.
They want explicit expression for V as a function of t, T as a function of t, and T as a function of V. So, you will need to go further than writing expressions for dV/dt etc.

b) I can't add together the work done in the volume change, then add work to change temperature, because of the path dependence of work. This is definitely an integral! Which I am really struggling to form...

For part (b) you are right that you will need to set up an integral. Your result for part (a) should be helpful.

 

[whatisreality]

For part (a) your expressions look OK except for a couple of sign errors. Note that the volume is decreasing while T and t increase.
They want explicit expression for V as a function of t, T as a function of t, and T as a function of V. So, you will need to go further than writing expressions for dV/dt etc.
For part (b) you are right that you will need to set up an integral. Your result for part (a) should be helpful.

So I need to integrate my expressions for dV/dt and dT/dt? Which should be dV/dt = -0.154 and dT/dt = 211.5. An indefinite integral, would it be?

 

[TSny]

You can use indefinite integrals. You will then need to determine the arbitrary constants of integration.

 

[whatisreality]

You can use indefinite integrals. You will then need to determine the arbitrary constants of integration.

Right. So dV=-0.154 dt, integrate to get V=-0.154t+c, and v=0.616 when t=0.
So V= -0.154t+0.616
dT = 211.5dt , so T=211.5t+c. T=300.15K when t=0 (I assume I have to convert to kelvin), so
T = 211.5t + 300.15
Then rearrange these expressions to get the dependence of T on V.

 

[TSny]

Yep. Good.

 

[whatisreality]

Yep. Good.

The answer to part a, then, is
(T-300.15) / 211.5 = (0.616 - V) / 0.154

300.15 + 211.5*(0.616 - V) / 0.154 = T
Still not sure what to actually integrate though. Pressure is changing too, since it doesn't state otherwise. I was expecting to integrate pressure with respect to volume.

 

[whatisreality]

Use pV=nRT to find an expression for pressure in terms of volume maybe?

 

[TSny]

Use pV=nRT to find an expression for pressure in terms of volume maybe?

Yes. That's it. (Your expression for T as a function of V can be simplified.)

 

[whatisreality]

Yes. That's it. (Your expression for T as a function of V can be simplified.)

Simplifies to 1146.15 - 211.15V/0.154 = T. Sub in:
p = nR(1146.15 / V) - 211.5nR/0.154
And n=30, R=8.31, but I think that just makes it more complicated. So leaving them as n and R and integrating dV gives

W = 1146.15nR ln(V) - 211.5nRV/0.154 + c, but I can do a definite integral between V₂ and V₁

 

[TSny]

Yes.

 

[whatisreality]

Yes.

Which means work is -32926.85...
And I probably have to use that to work out the next part too. Is this one the first law of thermodynamics? I don't have internal energy so maybe not. And heat is Q=mc dT, but I don't know what c is because the question justs says 'an ideal gas'.
Some sort of conservation of energy equation... Heat is energy in transit, isn't it?

 

[whatisreality]

No, not energy conservation, not a closed system, heat is added.

 

[TSny]

I don't agree with your numerical answer for the work.

You should know how the internal energy of a monatomic ideal gas depends on T.

 

[whatisreality]

I don't agree with your numerical answer for the work.

You should know how the internal energy of a monatomic ideal gas depends on T.

I'm a little confused. I know the equation
change in internal energy = nC_v dT, but Cv= 3R/2 is the heat capacity at constant volume... Except my textbook did a u-turn and says it applies whether volume is constant or not. So I can use that equation after all.

 

[TSny]

Right. Internal energy is a state function: U = nC_vT.
ΔU is independent of path.

 

[whatisreality]

Right. Internal energy is a state function: U = nC_vT.
ΔU is independent of path.

Oh! Ok, I remember that. And I don't agree with my previous numerical answer either... Now I have -92602.6448...
Since I can calculate the internal energy, Q = dU +W, so sub in T₁ and T₂ to work out dU. etc.

 

[TSny]

Yes. I agree with your new value for W. Sounds good!

 

[whatisreality]

Yes. I agree with your new value for W. Sounds good!

dU = 158180.85 and Q = 65578.2052. Provided I typed it in right!

Thank you for your time and help :) I was so confused!

 

[TSny]

OK. Of course, think about significant figures and units. Good Work!