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KCET 2025 Physics Paper

Question : 12 of 60

Marks: +1, -0

A gas is taken from state A to state B along two different paths 1 and 2. The heat absorbed and work done by the system along these two paths are

Q_1

and

Q_2

and

W_1

and

W_2

respectively. Then

$W_1=W_2$
$Q_1-W_1=Q_2-W_2$
$Q_1+W_1=Q_2+W_2$
$Q_1=Q_2$

Validate

Solution:

According to the first law of thermodynamics, the change in internal energy

(\Delta U)

of a system can be expressed as:

\Delta U=Q-W

Where

Q

is the heat absorbed by the system and

W

is the work done by the system.

For paths 1 and 2 from state

A

to state

B

, the change in internal energy is the same, regardless of the path taken:

\Delta U_1=\Delta U_2

Therefore, the heat absorbed and work done along the different paths follows the relationship:

Q_1-W_1=Q_2-W_2

This equation signifies that the net change in energy (considering both heat and work) is equal for both paths due to the same change in internal energy from state A to state B.

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