# Homework Help: Heat required to increase the temperature

1. Jul 27, 2017

### gj2

1. The problem statement, all variables and given/known data

At high temperatures the nitrogen molecule behaves like a one-dimensional harmonic oscillator. In this situation, estimate how much heat must be added to the system in order to increase the temperature of 1 mole of nitrogen gas by 10 degrees Celsius (for constant volume and constant pressure respectively). Take into account all degrees of freedom: translational, rotational and vibrational.

2. Relevant equations
Average energy per degree of freedom: $kT/2$

3. The attempt at a solution
A one-dimensional harmonic oscillator has two degrees of freedom, therefore according to the equipartition theorem the average energy of a nitrogen molecule must be $kT$. One mole of nitrogen has $N_A$ molecules and so the total internal energy of the gas is $U=N_A k T$. Therefore if the process is isochoric the amount of heat we need to add to the system in order to increase the temperature by 10 degrees is
$$dQ=N_A k dT=N_A k \cdot 10\text{K} \approx 19.8 \text{cal}$$
However the correct answer is $70 \text{cal}$ and I don't understand why.
Also, I have no idea how to find the heat for the case of an isobaric process since this is not an ideal gas (the answer for constant pressure case is $90 \text{cal}$).

2. Jul 27, 2017

### Orodruin

Staff Emeritus
First of all, a nitrogen molecule is not a one dimensional harmonic oscillator. You are missing a lot of degrees of freedom by this assumption.

3. Jul 27, 2017

### gj2

But that's how the question was formulated. It's not my assumption.

4. Jul 27, 2017

### Orodruin

Staff Emeritus
It is an oscillator that in addition can translate and rotate. Only the vibrational spectrum is described by an actual harmonic oscillator.

5. Jul 27, 2017

### gj2

Oh I see. So the translational motion contributes 3 degrees of freedom whereas the rotational contributes 2. Therefore there are 7 degrees of freedom and the heat required is approximately 70cal, as expected.
But what about the constant pressure case? How can I approach this problem?
Thank you.
Edit: nevermind. I think the reason why it's $90 \text{cal}$ for constant pressure case is because $C_p-C_v=2 \frac{\text{cal}}{\text{mol}\,\text{K}}$ for nitrogen (apparently it does behave approximately as an ideal gas). So it requires additional $$\left(2 \,\frac{\text{cal}}{\text{mol}\,\text{K}} \right )\cdot (10 \,\text{K})=20 \,\frac{\text{cal}}{\text{mol}}$$ heat for constant pressure case.

Last edited by a moderator: Jul 27, 2017