2 mol of an ideal gas undergoes a reversible and isothermal expansion from volume of 2.5 L
10 1 at 27°C. Calculate the work done by the gas in this expansion.
to 10 L to 10 L at 27°C. Calculat
Given R = 8.314 J/K/mol.
Answers
The work done by the gas in the isothermal and reversible expansion is -6.916 kJ.
• Work done by a gas undergoing a reversible and isothermal expansion is given by the formula,
W = -2.303 nRT log (V₂ / V₁) - (i)
where W is the work done by the gas,
- sign indicates that the work is done by the gas,
n is the number of moles of the gas,
R is the gas constant,
T is the temperautre of the gas,
V₂ the final volume of the gas,
and V₁ the initial volume of the gas.
• Given,
n = 2 moles
V₁ = 2.5 L
V₂ = 10 L
T = 27°C = (27 + 273 ) K = 300 K
R = 8.314 J / K / mol
• Substituting these values in equation (i) , we get,
W = -2.303 × 2 × 8.314 × 300 ×
log (10/2.5)
=> W = -11488.28 × log 4
=> W = -11488.28 × 0.6020
=> W = - 6915.94 J
=> W = -6915.94 × 10⁻³ kJ
=> W = - 6.916 kJ