Question

One mole of an ideal monoatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27c. The work done on the gas will be:

Answer 1

Hey mate,

◆ Answer - 300Rln(2)

◆ Explanation-
# Given -
n = 1
T = 27 °C = 300 K
P2 = 2P1

# Solution-
Work done on gas in isothermal process is calculated by-
W = nRT×ln(P2/P1)
W = 1×R×300×ln(2P1/P1)
W = 300RIn(2)

Work done on gas in doubling the pressure isothermally will be 300Rln(2).

Hope that is useful...