Question

Find the change in the internal energy of a gas when it absorbs 80 calories of heat and performs work equal to 170 joules

Answer 1

dQ=du+dw
80 cal = 334.7 J
By sign convention

dQ=+334.7J ---Heat absorbed is positive
dw=−170J --- heat given out is negative
du=dQ−dw Work done by the system is positive
=334.7−(−170) Work done on the system is negative
=504.7J