First slide

First law of thermodynamics

Question

In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas releases 40 joules of heat and 30 joules of work was done on the gas. If the initial internal energy of the gas was 20 joules, then the final internal energy will be

Easy

A

2 J
B

-18 J
C

10 J
D

58 J

Solution

$ΔQ = ΔU + ΔW = (U_{f} - U_{i}) + ΔW$
$\Rightarrow - 40 = (U_{f} - 20) - 30 \Rightarrow U_{f} = 10 J$

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