A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 ml to 375 ml at a constant temperature of 37°C. As it does so, it absorbs 208J of heat. The value of q and w for the process will be: (In 7.5 = 2.01)

  • Option 1)

    q = -208 J, w = +208 J

  • Option 2)

    q = 208 J, w = 208 J

  • Option 3)

    q = 208 J, w = -208 J

  • Option 4)

    q = -208 J, w = -208 J

 

Answers (1)

As we discussed in concept

First law of Thermodynamics -

Energy of universe is always conserved or total energy of an isolated system is always conserved

\Delta E= q + W


 

- wherein

\Delta E= Internal Energy

q= Heat

W= work

 

 Process is isothermal reversible expansion.

Therefore, \Delta E=0

Now using first law of thermodynamics.

\Delta E=\:g+W;g=208\:J\:(given)

W = -g

    = - 208 J


Option 1)

q = -208 J, w = +208 J

This option is incorrect.

Option 2)

q = 208 J, w = 208 J

This option is incorrect.

Option 3)

q = 208 J, w = -208 J

This option is correct.

Option 4)

q = -208 J, w = -208 J

This option is incorrect.

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Exams
Articles
Questions