When a system is taken from state i to state f along the path iaf, it is found that $\mathrm{Q}=50 \mathrm{~cal}$ and $\mathrm{W}=20 \mathrm{~cal}$. Along the path ibf $Q=36 \mathrm{~cal}$. W along the path ibf is

JEE Main · 2007 · 29 Apr 2007 · Offline · Heat & Thermodynamics

When a system is taken from state i to state f along the path iaf, it is found that

Q = 50 cal

and

W = 20 cal

. Along the path ibf

Q = 36 cal

. W along the path ibf is

(A)6 cal✓ Correct
(B)16 cal.
(C)66 cal.
(D)14 cal.

Answer: (A)

Solution

Internal energy is a state function, so the change

Δ U

between the states i and f is the same whatever path is followed. Along iaf,

Δ U = Q - W = 50 - 20 = 30

cal. Along ibf the same change must hold, so

36 - W = 30

, which gives the work along that path as

W = 6

cal.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

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