First slide

First law of thermodynamics

Question

A sample of an ideal gas is taken through a cycle a shown in figure. It absorbs 60J of energy during the process AB, no heat during BC, rejects 70J during CA. 40J of work is done on the gas during BC. Internal energy of gas at A is 1500J, the internal energy at C would be

Moderate

A

1590 J
B

1620 J
C

1540 J
D

1570 J

Solution

$\begin{array}{l} {ΔW}_{AB} = 0 as V = constant \\ ∴ {ΔQ}_{AB} = {ΔU}_{AB} = 60 J (Given) \\ U_{A} = 1500 J ∴ U_{B} = (1500 + 60) J = 1560 J \\ {ΔW}_{BC} = - {ΔU}_{BC} = - 40 J (Given) \\ ∴ {ΔU}_{BC} = 40 J ∴ U_{C} = (1550 + 40) J = 1590 J \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App