Question

A system absorbs 500J heat and utilized a part
in doing work at constant T. If change in internal
energy is -1005J, Work done is (in Itr-atm)
1) + 5 2) + 10 3) – 505 4) - 5
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Answer 1

Given:

A system absorbs 500J heat and utilized a part  in doing work at constant T. The change in internal  energy is -1005J.

To find:

Work done is (in Itr-atm)

Solution:

From given, we have,

A system absorbs 500J heat and utilized a part  in doing work at constant T.  

⇒ Δ Q = 500 J

The change in internal  energy is -1005J,

⇒ Δ U = -1005 J

we use the formula,

Δ U = Δ W - Δ Q

-1005 = Δ W - 500

-1005 + 500 = Δ W

∴ Δ W = -505 J

1 J = 1/101.3 ltr-atm

- 505 J = - 505/101.3 ltr-atm

∴ - 505 j = -4.98 ltr-atm ≈ -5 ltr-atm

Option  4) -5 is correct.