JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Heat & Thermodynamics

A thermodynamic system is taken from an initial state

i

with internal energy

U_{i} = 100 J

to the final state

f

along two different paths iaf and

ib f

, as schematically shown in the figure. The work done by the system along the paths

a f, ib

and

b f

are

W_{a f} = 200 J, W_{ib} = 50 J

and

W_{b f} =

100 J

respectively. The heat supplied to the system along the path

ia f, ib

and

b f

are

Q_{ia f}, Q_{ib}

and

Q_{b f}

respectively. If the internal energy of the system in the state

b

U_{b} = 200 J

and

Q_{ia f} =

500 J

, the ratio

Q_{b f} / Q_{ib}

Answer: 2

Solution

Path iaf:

Q_{ia f} = Δ U_{ia f} + W_{ia f}

. Here

W_{ia f} = W_{ia} + W_{a f} = 0 + 200 = 200

J (assuming

ia

is isochoric).

500 = Δ U_{ia f} + 200

, so

Δ U_{ia f} = U_{f} - U_{i} = 300

J, giving

U_{f} = 400

J.

Path ib:

Q_{ib} = Δ U_{ib} + W_{ib} = (U_{b} - U_{i}) + W_{ib} = (200 - 100) + 50 = 150

J.

Path bf:

Q_{b f} = Δ U_{b f} + W_{b f} = (U_{f} - U_{b}) + W_{b f} = (400 - 200) + 100 = 300

J.

Ratio:

\frac{Q _{b f}}{Q _{ib}} = \frac{300}{150} = 2

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer