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Grade 11 Science · Problem type

The first law of thermodynamics

Use this for “A gas takes in 500 J of heat and does 200 J of work pushing a piston out. How much does its internal energy change?”

Diagram

Q = 500 J inW = 200 J by the gaswarmer: faster500Q−200−W300ΔUenergy (J)
First law: ΔU = Q − W = 500 − 200 = 300 J
Q is + when heat flows into the gas; W is + when the gas does work by pushing the piston out.
More energy came in than went out as work: the internal energy rises and the gas warms.
This is an example. Type your own values to start a new problem.
ΔU (J)
=
Q (J)
−
W (J)

Formulas

ΔU
=
Q
−
W
300 = 500 − 200

Assumptions

•
Energy is kept: the heat in goes to the gas’s internal energy or out as work. ΔU = Q − W.
•
Q is + for heat into the gas and − for heat out of it.
•
W is + when the gas does work by expanding and − when work is done on it (it is squeezed).
•
Some books (AP Physics 2) count W as work done on the gas and write ΔU = Q + W.

Step-by-step

WE KNOW
Q = 500 J, W = 200 J
FIND
change in internal energy (ΔU)
Step 1 · Find change in internal energy (ΔU)
Use ΔU = Q − W
The heat in less the work out.
ΔU = 500 − 200
ΔU = 300 J
Check
Put the numbers back in. Both sides should match.
300 = 500 − 200 ✓
Topic check
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