Specific heat for an ideal gas

[grangr]

Homework Statement

If C_p for an ideal gas is 35.4 J/mol⋅K, which of the following is C_v for this gas?

a. 12.5 J/mol⋅K
b. 20.8 J/mol⋅K
c. 29.1 J/mol⋅K
d. 27.1 J/mol⋅K
e. 43.4 J/mol⋅K​

Homework Equations

1. ΔH = ΔE + Δ(PV) = Q + W + Δ(PV), and for ideal gas, ΔH = nC_vΔT + Δ(nRT) = nC_vΔT + nRΔT = nC_pΔT
2. The kinetic energy for a gas: KE = 3/2⋅(nRT)
3. From (1) and (2) above, for an ideal gas, C_v = 3/2⋅(R), thus C_p = 5/2⋅(R)

The Attempt at a Solution

C_p = 35.4 J/mol⋅K = 5/2⋅(R)
C_v = 3/2⋅(R) = 21.24 J/mol⋅K
The correct answer is (d). Clearly I am missing something here, and would appreciate your help!

 

[Orodruin]

Are there any internal degrees of freedom specified?

 

[grangr]

Are there any internal degrees of freedom specified?

No, I believe there is not. The problem statement was all that was given.

 

[ehild]

Homework Statement

If C_p for an ideal gas is 35.4 J/mol⋅K, which of the following is C_v for this gas?

a. 12.5 J/mol⋅K
b. 20.8 J/mol⋅K
c. 29.1 J/mol⋅K
d. 27.1 J/mol⋅K
e. 43.4 J/mol⋅K​

Homework Equations

1. ΔH = ΔE + Δ(PV) = Q + W + Δ(PV), and for ideal gas, ΔH = nC_vΔT + Δ(nRT) = nC_vΔT + nRΔT = nC_pΔT

Factor out nΔT from nC_vΔT + nRΔT: You get nΔT(C_v+R) = nΔT C_p . So how are C_p and C_v related?

 

[grangr]

Factor out nΔT from nC_vΔT + nRΔT: You get nΔT(C_v+R) = nΔT C_p . So how are C_p and C_v related?

In my attempt, I thought C_v = (3/2)R and C_p = (5/2)R, as reasoned below.

The kinetic energy (KE) for a gas is 3/2⋅(nRT) (Relevant equation #2). For an ideal gas, now that

1. ΔE = nC_vT, and
   
2. KE = 3/2⋅(nRT)

C_v = 3/2⋅R (Relevant equation #3)

Therefore, given that ΔH = n(C_v+R)ΔT = n([3/2⋅R]+R)ΔT = n(5/2)RΔT = nC_pΔT
C_p = (5/2)R (Relevant equation #3)

 

[Orodruin]

I thought Cv = (3/2)R

This is true only for a monoatomic gas without internal degrees of freedom. The gas constant R has a fixed value (around 8.31 J/(mol K)), it is not something you solve for. Hence my hint regarding internal degrees of freedom and @ehild's hint regarding the relation between $C_p$ and $C_v$.

 

[grangr]

This is true only for a monoatomic gas without internal degrees of freedom. The gas constant R has a fixed value (around 8.31 J/(mol K)), it is not something you solve for. Hence my hint regarding internal degrees of freedom and @ehild's hint regarding the relation between $C_p$ and $C_v$.

I get it. For ideal gas, ΔH = nC_vΔT + Δ(nRT) = nC_vΔT + nRΔT = nC_pΔT, thus C_v + R = C_p. Therefore, C_v = C_p - R = 35.4 - 8.314 [J/mol⋅K] = 27.086.

To be honest though, I fail to comprehend the concept of internal degrees of freedom still. The question asks about an ideal gas, supposedly it is not wrong to assume it is a monotomic-molecule gas (No?), in which case there would be 3 translational degrees of freedom and nothing else. So, that would give a kinetic energy (or internal energy) of (3/2)RT... (I am sorry; I don't think I am making any sense here. This is probably beyond my education.)

Thank you for all your replies.

 

[Orodruin]

The question asks about an ideal gas, supposedly it is not wrong to assume it is a monotomic-molecule gas (No?), in which case there would be 3 translational degrees of freedom and nothing else.

Yes, it is wrong. If it had no internal degrees of freedom, then it would have $C_p = 5R/2$, which is not the case. If nothing is mentioned about the gas being monoatomic or not it is not something you should assume. Of course, if you would have been given a value lower than 5R/2 you should have started becoming suspicious. From the data in the problem, you could deduce that the gas was not monoatomic. An ideal gas generally is not required to be monoatomic.

 

[ehild]

I get it. For ideal gas, ΔH = nC_vΔT + Δ(nRT) = nC_vΔT + nRΔT = nC_pΔT, thus C_v + R = C_p. Therefore, C_v = C_p - R = 35.4 - 8.314 [J/mol⋅K] = 27.086.

Correct.

To be honest though, I fail to comprehend the concept of internal degrees of freedom still. The question asks about an ideal gas, supposedly it is not wrong to assume it is a monotomic-molecule gas (No?), .

No, you cannot assume it is monoatomic gas. It can be two-atomic like O2 and N2, or three atomic as CO2 or even more-atomic or the mixture of to different substances. But Cp = Cv +R is substantial law for ideal gases.

 

[mjc123]

Yes, it is wrong to assume it is a monatomic gas. There is no reason why a polyatomic gas cannot be ideal, as long as it obeys PV = nRT. C_v will be 3R/2 + R/2 for each active rotational degree of freedom and R for each active vibrational degree of freedom. The important thing you have to remember, and the only thing you really needed to know for this question, is that for an ideal gas C_p = C_v + R under all circumstances.

 

[Orodruin]

I fail to comprehend the concept of internal degrees of freedom still.

So, that would give a kinetic energy (or internal energy) of (3/2)RT...

To clarify a bit. The point is that if you have N degrees of freedom, you will have an internal energy of NRT/2. If you have a monoatomic gas, then the three kinetic degrees of freedom are the only ones you have. If you have more degrees of freedom, then you will need to add more heat to change the temperature by the same amount, i.e., a higher heat capacity.

Edit: Based on #10, let me clarify that the double counting of the vibrational degrees of freedom arise from the corresponding potential. In the case of translation/rotation of the molecule there is no corresponding potential.

Edit 2: Of course, the temperature also needs to be large enough to excite the vibrational degrees of freedom for them to have an effect.