# Both molar specific heat of an ideal monatomic gas at constant volume and pressure

1 vote
70 views

2 mols of oxygen are heated up from a temperature of 20 C and a pressure 1 atm to a temperature of 100 C. Compute:
a) Having constant volume, how much heat must be provided to the gas throughout the process?
b) Having constant pressure, how much work is done throughout the process?
c) Compute in a) and b) cases the variation in the internal energy.

My try:

The molar specific heat of an ideal gas at constant volume is:

$$C_v = \frac{3}{2}R$$

The molar specific heat of an ideal gas at constant pressure is:

$$C_p = \frac{5}{2}R$$

Then at a) it is just about using $Q = nC\Delta T$. My issue here is that the given outcome at a) uses n = 1:

$$Q = nC_v\Delta T = 997.68J$$

I obtained 1995.36J (I used n=2).

To compute Q at constant pressure:

$$Q = nC_p\Delta T = 1662.80J$$

I obtained 3325.6J (I used n=2).

I do not understand why it uses n=1 as it is specified that the gas is formed by 2 mols of oxygen.

b) When heat is supplied at constant pressure the gas disseminates and exerts a positive work (such as on a piston).

However the answer is given as a negative number, which means work is done by the system and not over it. Thus the given answer:

$$W = -nR\Delta T = -1330.24J$$

Here I got the same but positive.

c) Using the first principle of thermodynamics:

$$\Delta E_i = Q + W$$

When volume is constant:

$$\Delta E_i = Q_v + W = -332.56J$$

When pressure is constant:

$$\Delta E_i = Q_p + W = 332.56J$$

What I got before seeing the answers:

When volume is constant:

$$\Delta E_i = Q_v + W = 3325.60J$$

When pressure is constant:

$$\Delta E_i = Q_p + W = 4655.84J$$

As you can see all comes down to the definition: $Q = nC_v\Delta T$ I have made some research and in Tipler and $C_v$, $C_p$ and Q are defined:

$$C_v = n\frac{3}{2}R$$

$$C_p = n\frac{5}{2}R$$

$$Q = C\Delta T$$

Eventually it is the same, since I multiplied by 2 (number of mols) in the definition: $Q = nC\Delta T$ and I did not do it my first $C_v$ and $C_p$ definitions. Therefore I think my outcomes are right but what do you think?

asked Aug 9, 2018
edited Aug 10, 2018
Sorry I do not understand your question ("My issue here is..."). Are you saying that the answers which are given in your textbook are correct for 1 mol of gas instead of 2 mols? Please give more explanation about your difficulty.
Okey I updated my question
I cannot see a reason for it. What are the answers to parts (b) and (c)? Do they make the same mistake?
Updated. What do you think about the issue?
I agree with your answer to (b). Energy of 3326J is supplied to the gas, this increases the internal energy of the gas by 1995J while the gas does work of 1330J one the atmosphere. The gas does +ve work on the atmosphere, the atmosphere does -ve work on the gas.

In (c) the internal energy of the gas is increased by the same amount (1995J) in both cases, because internal energy depends only on temperature and number of moles, and the increase in temperature is the same in both cases.
OK I know what you mean at c). I have read that if all internal energy is regarded as transitional kinetic energy the internal energy equals to $\frac{3}{2}nRT$ . But I do not understand why there is not potential energy, I mean, attractive force between particles is conservative and therefore there has to be a potential energy definition. I guess it can be cancel out, but how?
Your title says that the gas is "ideal monatomic". This means that there are no interactions between molecules, and no interactions between atoms inside each molecule (because the "mono-atom" does not interact with itself!). Therefore there is no potential energy stored in the gas.