# Homework Help: 1st Law of Thermodynamics question

1. Oct 30, 2015

### Jamessamuel

1. The problem statement, all variables and given/known data
A rigid vessel contains 0.1kg of air initially at 1bar and 20degC. A heat transfer to the gas
increases the temperature to 1000degC. What type of process does the gas undergo ? Calculate
(i) the initial and final volume of the air
(ii) the final pressure
(iii) the change in internal energy and enthalpy
(iv) the work and heat transfers

2. Relevant equations
1st law eqn

3. The attempt at a solution
i) pV = nRT using T = 293.15K and finding n using the molar air mass, p is given as 1 bar.
solution is around 0.084 m^3 (2sf) , (V = nRT/p)

ii) constant vol, so p/T = const. using the initial and final states, finding the 2nd pressure results from some simple calculations i.e p(2) = p(1)/T(1) * T(2).
solution is aorund 4.4bar. so far so good.

iii) this is where the books' answer and my own are different.
My solution:
change in internal energy = mass * specific heat at constant vol * temp change
mass = 0.1kg
temp change = 980
SH =1.5R where R = univ. gas constant / molar mass(air) = 8.3145/0.02897
giving SH = 430.5 (4sf)
hence change in internal energy = 42 kJ(2sf) (multiplying the three terms)
The book gives 70.6kJ?
for the enthalpy, continuing with my solution, it is = change in internal energy + any work done
but no work is done, so the enthalpy = the heat transfer is the change in internal energy (const.volume)
hence equals 42kJ.
The book gives 99kJ?

iv) i said no work done, so did book. the book said the heat transfer is 70.6kJ? i would have said my 42kJ.

also, despite the question being in the 1st law part, i thought that only applied at constant pressure? so can't i apply it here?

i appreciate im asking a lot, any help appreciated.

Regards,

James.

2. Oct 30, 2015

### TSny

Did you take into account that air is essentially a diatomic gas?

Change in enthalpy is not always equal to the change in internal energy plus the work done. Recall the definition of enthalpy in terms of internal energy, pressure, and volume.

The first law is a general law of thermodynamics. It is not restricted to constant pressure processes.

3. Oct 31, 2015

### Jamessamuel

do you agree with my internal energy solution? (iii part 1), or does the diatomic property influence the solution method?

4. Oct 31, 2015

### TSny

In your solution to (iii) you did not use the correct molar heat capacity for a diatomic gas at constant volume. It is not 1.5R.

5. Oct 31, 2015

### Jamessamuel

ok, i did some research;

for a monatomic gas the molecules only have 3 degrees of freedom corresponding to translational kinetic energy in 3 dimensions. But for a diatomic gas, there are 5- 2 extra degrees of rotational freedom. Am i right in assuming that air is a composition of diatomic molecules, and that my appended molar heat capacity should not be 1.5R, but 2.5R?

6. Oct 31, 2015

### Jamessamuel

my way of justifying the idea that each degree of freedom corresponds to 0.5R of the energy is that we assume that the gas is isotropic, hence the contributions from each dimension are equally weighted.

7. Oct 31, 2015

### TSny

Yes, 2.5R. This ignores the vibrational degree of freedom of nitrogen and oxygen http://scied.ucar.edu/molecular-vibration-modes. The vibrational modes can be ignored at room temperature. In this problem the temperature goes to 1000 oC. I'm not sure if it's safe to ignore the vibrational mode at this temperature. But, anyway, I think you will get the answers provided if you use 2.5R for the molar heat capacity at constant volume.

8. Oct 31, 2015

### TSny

I found these tables:
https://www.ohio.edu/mechanical/thermo/property_tables/air/air_Cp_Cv.html.
http://www.nist.gov/data/PDFfiles/jpcrd581.pdf [Broken] (page 367)

These tables show that the heat capacity at constant volume for air increases by about 25% as the temperature increases from 20 oC to 1000 oC. But I don't think you are meant to worry about this in your problem.

Last edited by a moderator: May 7, 2017