First slide

Change in internal energy

Question

In the figure given two processes A and B are shown by which a thermo-dynamical system goes from initial to final state F. If ΔQ_A and ΔQ_B are respectively the heats supplied to the system then

Easy

A

$\Delta {Q_A} = \Delta {Q_B}$
B

$\Delta {Q_A} \geqslant \Delta {Q_B}$
C

$\Delta {Q_A} < \Delta {Q_B}$
D

$\Delta {Q_A} > \Delta {Q_B}$

Solution

The area under the graph 'iAf' is greater than area under the graph 'iBf'
So work done by teh gas in process A is greater than the work done by the gas is process A is greater than that in process B.
Also ΔU = (U_f - U_i) is same for both process as internal energy is a state function.

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