First slide

First law of thermodynamics

Question

A perfect gas goes from state A to another state B by absorbing 8X10⁵ J of heat and doing 6X10⁵J of external work. It is now transferred between the same two states in an other process in which it absorbs 10⁵ J of heat. In the second process

Easy

A

work done by gas is 10⁵ J
B

work done on gas is 10⁵ J
C

work done by gas is 0.5 X10⁵ J
D

work done on the gas is 0.5x 10⁵ J

Solution

In the first process
$ΔU = (8 \times 10^{5} - 6 \times 10^{5}) J = 2 \times 10^{5} J$
For the second process
$Q^{'} = ΔU + W^{'} \Rightarrow W^{'} = (10^{5} - 2 \times 10^{5}) J = - 10^{5} J$

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