# Homework Help: Heat, Work, Change in Entropy and Energy

1. Nov 2, 2017

### ScreamingIntoTheVoid

1. The problem statement, all variables and given/known data
Calculate q, w, ∆E, and ∆H for the process in which 93.0 g of nitrous oxide gas (N2O) is cooled from 179°C to 55°C at a constant pressure of 4.00 atm.

Cp(N2O) = 38.70 J K-1 mol-1

2. Relevant equations
q= mCΔT
ΔH=n(Cp)=n(qv)ΔT
ΔE=q+w
w= -pΔV
*Probably something else too but I'm drawing a blank
3. The attempt at a solution
q= (93.0g)(38.70 K-1Mol-1)(-124°C) → -446288.4
w= -(4 atm) (101.3) (?) = ??? *not 0 though*
ΔE=-446288.4+w=???
ΔH= (2.113 moles)(-446288.4)(-124°C)= A really big and incorrect number

Help?

2. Nov 2, 2017

### Staff: Mentor

In your calculation of q, the units are incorrect. The grams and moles don't cancel.

From the ideal gas law, what is the initial volume? What is the final volume?

Your equation for $\Delta H$ is incorrect.

3. Nov 2, 2017

### ScreamingIntoTheVoid

Ok, in that case, q=nCp delta T -> (2.113 moles)(38.70)(-124) = -10139.8644 J
For the volumes I used pv=nRT, in which I calculated 19.55... for V1 and 14.175..., leaving the difference to be approximately 5.3752...
w= p delta V -> (4 x 101.3) (5.3752 L)= 2178.033824 J. Since work was put into the system to compress it, work is positive right?
Delta E= 2178.033824 + -10139.8644 J= -7961.830576 J?

Is any of that correct?
What formula should I use for Entropy instead?

4. Nov 2, 2017

### Staff: Mentor

Correct. This is also equal to $\Delta H$
Correct
Correct
Correct, although I don't like how you "wild assed" this.

Instead of what? What equations do you know for calculating the change in entropy for an ideal gas?

5. Nov 2, 2017

### ScreamingIntoTheVoid

The one I put above was the one I had in my notes. DeltaH= qp was also in there, but I looked over it I suppose.

"Wild assed ay? Is there a better way to calculate this to ensure accuracy (I have a midterm tomorrow).

THANK YOU very much!!

6. Nov 2, 2017

### Staff: Mentor

$\Delta H$ is not the change in entropy. It is the change in enthalpy. Entropy has the symbol S.

I would have done the other part this way:

$$W=-P\Delta V=-(4)(101.3)(-5.3752)=+2178 J$$

$$\Delta (PV)=-2178 J$$

$$\Delta H=\Delta E+\Delta (PV)$$
so
$$\Delta E=\Delta H-\Delta (PV)=-10140-(-2178)=-10140+2178=-7962 J$$