Question & Answer: A balloon is heated by adding 1000 J of heat. It expands by about 5 liters against a constant…..

Ans. Given-

I. 1000 J heat is supplied to the system (balloon).

So, Q = + 1000 J

II. Balloon expands by a volume of 5.0 L.

So, work done by the system during expansion is given by-

W =- p dV

where, p = 1 atm = 1.01325 x 10⁵ Pa

dV = 5.0 L = 5.0 x (10^-3 m³) = 5.0 x 10^-3 m³

Or, W = (1.01325 x 10⁵ Pa) x (5.0 x 10^-3 m³)

Or, W = 506.625 Pa m³

Or, W = 506.625 J                                       ; [1 J = 1 Pa m³]

Note that, work is done by the system on surroundings.

Now, Change is internal energy of the system is given by first law of thermodynamics as follow-

dU = Q – W

where, dU = Change in internal energy

Q = Heat supplied to the system

W = Work done by the system.

Pitting the values from I and II in above equation-

dU = 1000 J – 506.625 J

Hence, dU = 493.375 J = + 493.375 J

A positive value of dU indicates that there is net gain of heat by the system during the overall process (heat supply and expansion).