# Constant Volume Calorimetry - Why does (ΔnRT)=(Δn)RT

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1. Nov 30, 2015

### blithe285

When doing constant volume calorimetry, the enthalpy can be calculated as follows:

ΔH = ΔU + Δ(PV)
ΔH = w + q + Δ(PV)
ΔH = PΔV + q + Δ(PV)

and at constant volume:

ΔH = q + VΔP

which I've then see people rewrite using the ideal gas law as follows:

ΔH = q + (Δn)RT

where Δn is the change in the moles of gas and T is constant.

This is what I don't understand. Why is T constant? If you're doing calorimetry, the temperature is changing. Why are we now assuming that it is constant? Before looking it up, I originally had the following:

ΔH = q + nRΔT

Why isn't it this? Or even ΔH = q + ΔnRΔT

How do you know when to keep moles constant and when to keep the temperature constant?

Thanks!

2. Dec 1, 2015

### tommyxu3

Observe $PV=nRT,$ then $\Delta(PV)=\Delta(nRT)$ (just get the change of both sides). Since $V$ is constant, then $\Delta(PV)=(\Delta P)V.$
On the other side, $\Delta(nRT)=R\Delta(nT)=R(n\Delta T+T\Delta n),$ which you can check by $(n+\Delta n)(T+\Delta T)-nT.$ So the result must be modified, or he may had added the condition that $\Delta T$ can be ignored.

3. Dec 1, 2015

### DrDu

In a Calorimeter, you keep temperature constant. For example, you could measure the amount of ice which is molten with the heat supplied by the chemical reaction you are studying.

4. Dec 3, 2015

### blithe285

Mm. So there is also a change in temperature?

We had a homework problem dealing with constant volume calorimetry, and there was a temperature change of 2.5 K and a mole-of-gas change of -1/2. However, in the solutions, the equation they used was just

ΔH = q + (Δn)RT

And they just used the initial temperature of 25°C for T

Was this a mistake in the solutions? It makes so much for sense to me to use Δn and ΔT.

Thank you so much for your detailed response! It helped so much!

5. Dec 3, 2015

### blithe285

Thanks for the reply!

However, if you are doing a combustion reaction in a bomb calorimeter and it is allowed to exchange heat energy with a bath, you can measure the temperature change of the bath, right?

Hmm, actually now that I type it out, is it because the temperature of the stuff inside the bomb calorimeter isn't changing.. It's just exchanging heat with the bath, which IS undergoing a temperature change?

And then you can use the temperature change of the bath to find q, and then the (Δn)RT is an artifact of purely the system in the bomb calorimeter (which is then constant tamperature because it is exchanging heat with the bath to stay at constant T).

Is that correct?

Thank you so much!!

6. Dec 3, 2015

### blithe285

Actually but.. If the bath has a measurable temperature change because of heat exchange.. Wouldn't the system in the bomb calorimeter also have a temperature change via thermal equilibrium?? (0th Law)

Ugh, I'm confusing myself even more! >_<

7. Dec 4, 2015

### DrDu

You are completely right. I suppose, the neglect of the temperature change is an approximation.
So try to work out a more correct solution for youself.

Last edited: Dec 4, 2015
8. Dec 6, 2015

### James Pelezo

Would you mind defining what you mean by constant volume? That is, are you studying reactions in solution where a temperature change can be measured but the volume of the reaction solution does not change? (Example: Acid + Base => Heat in solution, measure temp change of sol'n and calculate enthalpy?)

9. Dec 8, 2015

### Staff: Mentor

In the bomb calorimeter, you are trying to measure ΔU for the reaction at constant volume and temperature. To hold the temperature constant, heat transfers out of the reaction chamber to the surrounding material of the calorimeter. The surrounding material has a (quantitatively known) much higher capacity to absorb heat so that, even though the temperature of the reaction mixture rises a little, the energy involved is insignificant compared to the amount of heat transferred to the surrounding material. The temperature rise of the surrounding material tells you how much heat had to be removed from the reaction chamber to maintain the temperature of the reaction mixture constant. So, from the first law, ΔU=Q, where ΔU is the change in internal energy in transitioning from the reactants to the products at constant temperature and volume. Usually, however, we are interested in ΔH for the reaction at constant temperature and pressure, rather than constant volume. For a reaction of ideal gases, we can get ΔH at constant temperature and volume by using ΔH=ΔU+Δ(PV)=ΔU+(RT)Δn, where Δn is the change in the number of moles of gas as a result of the reaction. But we want ΔH at constant temperature and pressure, not constant temperature and volume. Fortunately, we know that, for an ideal gas, enthalpy is a function only of temperature, and not pressure or volume. So, if we were to expand the final mixture in the reaction chamber back to the original pressure at constant temperature, the change in enthalpy for this operation would be zero. So, for ideal gas reactions, the change in enthalpy we obtain in the bomb calorimeter at constant temperature and volume is also the change in enthalpy at constant temperature and pressure.

10. Dec 8, 2015

### DrDu

But in a real world bomb calorimeter, you determine the ammount of heat transferred to the bath by its change of temperature. Hence they aren't operated at constant temperature.

11. Dec 8, 2015

### Staff: Mentor

Yes, I mentioned that the temperature of the reaction mixture changes a little. But, the thermal capacity of the reaction mixture is typically very small compared to the thermal capacity of the calorimeter, so, to a first approximation the change in internal energy for constant temperature is very close to the heat transferred to the calorimeter (bath). In any event, it is very easy to correct the result if you know the constant volume heat capacities of the product gases.