Specific heat capacity and changing volume

[DannyMoretz]

Hello everyone,
I just need some help understanding some thermodynamics. So I have 0.25 kg of helium which is compressed from an initial state in a polytropic process with n = 1.3. So its given the change in volume and the initial pressure. I need to find the change in internal energy. I am aware of the relationship ΔU = m . Cv . ΔT ... and I know that Cv = 3R/2, but can I use that particular internal energy equation, considering Cv is the specific heat capacity of helium at a constant volume, even though volume changes in this process ? Am I just misinterpreting the meaning of this ?

Thanks :)

 

[Delta2]

You should use that in a polytropic process the energy transfer ratio K=dQ/dW is constant and it is $n=(1-\gamma)K+\gamma$

 

[DannyMoretz]

You should use that in a polytropic process the energy transfer ratio K=dQ/dW is constant and it is $n=(1-\gamma)K+\gamma$

Thanks for the response, but I still don't understand how this answers my question.

 

[Delta2]

You can calculate the work done W and it will be Q=KxW. Now you know Q and W, its easy to find ΔU isn't it?

 

[DannyMoretz]

You can calculate the work done W and it will be Q=KxW. Now you know Q and W, its easy to find ΔU isn't it?

Yeah I know that part, but can I use that equation for change in internal energy ? It uses Cv, the specific heat capacity for a constant volume, but volume changes in this process ?

 

[Delta2]

No you can't use that equation because it is for isochoric processes and your process isn't isochoric (need n=infinite for a polytropic process to be isochoric)

 

[DannyMoretz]

No you can't use that equation because it is for isochoric processes and your process isn't isochoric (need n=infinite for a polytropic process to be isochoric)

Ok, thanks alot, I will explore the method you suggested.

 

[Chestermiller]

No you can't use that equation because it is for isochoric processes and your process isn't isochoric (need n=infinite for a polytropic process to be isochoric)

This is not correct. The molar heat capacity at constant volume is a physical property of a gas, defined by:
$$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$
where U is the internal energy per mole. In general, U = U(T,V), where V is the molar volume, so
$$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$
But, the internal energy of an ideal gas is independent of its specific volume. So, in general, for an ideal gas
$$dU=C_vdT$$
irrespective of whether the volume of the gas is constant.

Chet

 

[Delta2]

Ok i see you are right (well also we know that the change in internal energy for reversible processes depends only on the initial and final state not on the process itself). Still if he follows my approach he should get the same result .

 

[Chestermiller]

The easiest way to get the temperature change is use the ideal gas law: $nRΔT=Δ(PV)$. Once you know this, you can get the change in internal energy. Also, from the polytropic relationships, you get the work W. So, from all this you can then get the amount of heat Q.

Chet

 

[Chestermiller]

(well also we know that the change in internal energy for reversible processes depends only on the initial and final state not on the process itself).

We know that this is true even for irreversible processes.

Chet

 

[DannyMoretz]

This is not correct. The molar heat capacity at constant volume is a physical property of a gas, defined by:
$$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$
where U is the internal energy per mole. In general, U = U(T,V), where V is the molar volume, so
$$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$
But, the internal energy of an ideal gas is independent of its specific volume. So, in general, for an ideal gas
$$dU=C_vdT$$
irrespective of whether the volume of the gas is constant.

Chet

The easiest way to get the temperature change is use the ideal gas law: $nRΔT=Δ(PV)$. Once you know this, you can get the change in internal energy. Also, from the polytropic relationships, you get the work W. So, from all this you can then get the amount of heat Q.

Chet

Thanks a lot Chestermiller, this has fixed my understanding. I was about to tell my lecturer that he was doing something wrong :S.