Question

An ideal gas is cooled from 127 c to 27 c, the percentage change in internal energy of gas will be

Answer 1

Answer:

T1 = 927 + 273 = 1200 K

T2 = 27 + 273 = 300 K

internal energy ∝ Temperature

U ∝ T

(U1 / U2) = (T1 / T2)

∴ [(U1 – U2) / U2] = [(T1 – T2) / T2]

∴ % change in energy = [{(1200 – 300) / (300)} × (100)]

= 3 × 100 = 300%