The internal energy of a monoatomic gas is 3nRT. One mole of helium is kept in a cylinder having internal cross section area of 17 cm2 and fitted with a light movable frictionless piston. The gas is heated slowly by supplying 126 J heat. If the temperature rises by 4 , then the piston will move cm. (atmospheric pressure = 105 Pa)
- 1.55
- 14.5
- 15.5
- 1.45
Answer
(3) 15.5
Solution
Given U = 3nRT U = 3nR T = 3 1 8.3 4 = 99.6 J
First law: W = Q - U = 126 - 99.6 = 26.4 J
Light frictionless piston the gas expands at constant pressure P0 = 105 Pa:
W = P0 A x
x = 26.4105 17 10-4 = 26.4170 m
x 0.155 m = 15.5 cm
Answer: (3)