NEET-XII-Physics
26: Laws of Thermodynamics
- #15A gas is enclosed in a cylindrical vessel fitted with a frictionless piston. The gas is slowly heated for some time. During the process, 10 J of heat is supplied and the piston is found to move out 10 cm. Find the increase in the internal energy of the gas. The area of cross section of the cylinder = 4 cm2 and the atmospheric pressure = 100 kPa.Ans : Given: Heat supplied to the system, ∆Q = 10 J
Change in volume of the system, ∆V = Area of cross section × Displacement of the piston
= A × 10 cm
= (4 × 10) cm3 = 40 × 10-6 m3
P = 100 kPa
∆W = P∆V = 100 × 103 × 40 × 10-6 m3
= 4 J
∆U = ?
Using the first law of thermodynamics, we get
10 = ∆U + ∆W
⇒ 10 = ∆U + 4
⇒ ∆U = 6 J
Here, positive sign indicates that the internal energy of the system has increased.
Page No 63: