Question

A gas performs 0.320 kJ work on surrounding and absorbs 120 J of heat from the surrounding. Hence, change in internal energy is ______.

Options

• 200 J

• 120.32 J

• - 200 J

• 440 J

Answer 1

A gas performs 0.320 kJ work on surrounding and absorbs 120 J of heat from the surrounding. Hence, change in internal energy is - 200 J.

Explanation:

According to first law of thermodynamics,

Δ U = q + W

Since, work is done on the surrounding.

So, W = - 0.320 kJ

= - 320 J

q = 120 J

Δ U = - 320 + 120

= - 200 J