First slide

First law of thermodynamics

Question

A system is taken from state a to state c by two paths adc and abc as shown in the figure. The internal energy at a is $U_{a} = 10 J$ . Along the path adc the amount of heat heat absorbed $δ Q_{1} = 50 J$ and the work obtained $δ W_{1} = 20 J$ whereas along the path abc the heat absorbed $δ Q_{2} = 36 J$ . The amount of work along the path abc is

Moderate

A

10 J
B

12 J
C

36 J
D

6 J

Solution

According to first law of thermodynamics,
$δ Q = δ U + δ W$
Along the path adc
Change in internal energy,
$δ U_{1} = δ Q_{1} - δ W_{1} = 50 J - 20 J = 30 J$
Along the path abc Change in internal energy,
$δ U_{2} = δ Q_{2} - δ W_{2} δ U_{2} = 36 J - δ W_{2}$
As change in internal energy is path independent.
$∴ δ U_{1} = δ U_{2} ∴ 30 J = 36 J - δ W_{2} δ W_{2} = 36 J - 30 J = 6 J$

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