Answer:
-14892.93 J
Explanation:
given,
Pressure, P = 54700 Pa
heat removed, Q = 16800 J
radius, r = 0.272 m
distance, d = 0.150 m
internal energy ΔU of the system = ?
We know,
Force = Pressure x area
F = P x A
F = 54700 x π x r²
F = 54700 x π x 0.272²
F = 12713.79 N
Work done = F.d
W = 12713.79 x 0.15
W = 1907.07 J
Change in internal energy ΔU is
ΔU = W + Q
= 1907.07 + (-16800)
= -14892.93 J
Hence, the change in internal energy is equal to -14892.93 J