In: Physics
A cylinder, which is in a horizontal position, contains an unknown noble gas at 48400 Pa and is sealed with a massless piston. The piston is slowly, isobarically moved inward 18.9 cm, which removes 18600 J of heat from the gas. If the piston has a radius of 27.2 cm, calculate the change in the internal energy of the system Δ U
Solution :
Given :
P = 48400 Pa
L = 18.9 cm = 0.189 m
Q = -18600 J
And, r = 27.2 cm = 0.272 m
Here, Work done by the gas in isobaric process (Because Pressure inside the cylinder remains constant)
W = P V = P( - A L) = - P (r2) L = -( 48400 Pa)()(0.272 m)2(0.189 m) = - 2126.15 J
Now According to the first law of thermodynamics :
Q = W + U
U = Q - W
U = ( - 18600 J) - ( - 2126.15 J)
U = -16473.845 J = - 16474 J