How Is Work Related to Heat in Thermodynamic Processes?

[jagged06]

Homework Statement

One mole of a dilute gas initially has a pressure equal to 1.00 atm, and a volume equal to 12.0 L. As the gas is slowly heated, the plot of its state on a PV diagram moves in a straight line to the final state. The gas now has a pressure equal to 5.00 atm, and a volume equal to 76.0 L.

(a) Find the work done by the gas.
19.2 kJ [correct]

(b) Find the heat absorbed by the gas.
I'm just not sure if I would add Internal Energy and the Work, or subtract the work

Homework Equations

Δ U = Q + W

The Attempt at a Solution

E = (3/2) nR(T2-T1)
T1= (P1V1)/nR
T2=(P2V2)/nR

T1=((1x10^5)(.012))/R
T2=((5x10^5)(.076))/R

Here is where I'm not so sure:

is it Q= E-W
-> Q= 36 kJ

or Q= E+W
-> Q=74.4 kJ

 

[rock.freak667]

Work done by a gas is positive, so you'd add it.

 

[jagged06]

Correct, Thank you very much!