First slide

First law of thermodynamics

Question

A perfect gas goes from state A to state B by absorbing 8 X $10^{5}$ J of heat and doing 6.5 x 10⁵ J of external work. It is now transferred between the same two states in another process in which it absorbs 10⁵ J of heat. In the second process

Moderate

A

work done on the gas is 10⁵ J
B

work done on the gas is 0.5 X 10⁵J
C

work done by the gas is 10⁵ J
D

work done by the gas is 0.5 X 10⁵ J

Solution

According to first law of thermodynamics,
$d U = d Q - d W$
$d U = 8 \times 10^{5} - 6.5 \times 10^{5} = 1.5 \times 10^{5} J$
In the second process, dU remains the same.
$∴ d W = d Q - d U = 1 \times 10^{5} - 1.5 \times 10^{5}$
$d W = - 0.5 \times 10^{5} J$
$∴$ Work done on the gas is $0.5 \times 10^{5} J$

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