By how much does the thermal energy of the gas change?

[mssca]

This is my very first post... :) I found some help here: https://www.physicsforums.com/showthread.php?t=277092 But I have a difficulty in sloving second and third part

Homework Statement

The figure shows(check the link) a thermodynamic process followed by 1.90×10^−2 of hydrogen (link to the graph: http://tiny.cc/c9m0h
http://session.masteringphysics.com/problemAsset/1074025/3/knight_Figure_17_65.jpg

I get -50.7J for the work done on the gas which is right... now they want to calculate:

By how much does the thermal energy of the gas change?
AND
How much heat energy is transferred to the gas?

Please help me... I spent hours reading the textbook, but I still don't get it.

Homework Equations

Eth=W+Q
pV=nRT
Q=nCdeltaT

The Attempt at a Solution

By how much does the thermal energy of the gas change? After using above equations and
I get dalta T = 64.15859142 AND I get Eth=-15

For this question "How much heat energy is transferred to the gas?" the heat energy is what?

Please help me... Thank you

 

[Mindscrape]

You also know that a monotonic gas, such as hydrogen, has a change in internal energy as such:

∆E=3/2 n R ∆T

You can use some ideal gas law and some other reasoning to change this to an expression with PV. Once you know the energy change, the heat should be easy. Give it a try. :)

 

[mssca]

You also know that a monotonic gas, such as hydrogen, has a change in internal energy as such:

∆E=3/2 n R ∆T

You can use some ideal gas law and some other reasoning to change this to an expression with PV. Once you know the energy change, the heat should be easy. Give it a try. :)

I got it! Except for the last section "How much heat energy is transferred to the gas?" Is is -25.8 oR +25.8...? My dad thinks it should be +25.8. I have the answer for the last section but I don't know if it is +/-. How do you determine +/-? For part B I entered 24.9J but it came out wrong and when I entered -24.9J it came out right.

Thanks.

 

[Mindscrape]

Fortunately, heat has only one convention, which is heat added to the system.

∆U=Q+W (work done on the system)

-24.9=Q-50.7

Q=25.8

So, the gas did a certain amount of work in expanding (work was done by the gas), the internal energy decreased, and heat was added to the system.

 

[mssca]

Fortunately, heat has only one convention, which is heat added to the system.

∆U=Q+W (work done on the system)

-24.9=Q-50.7

Q=25.8

So, the gas did a certain amount of work in expanding (work was done by the gas), the internal energy decreased, and heat was added to the system.

Got it! THANK YOU SOOOO MUCH!