Question

A 1.5-mole sample of an ideal gas is gently cooled at constant temperature 320 K. It contracts

from initial volume 19 L to final volume V2. A total of 1.2 kJ of heat is removed from the gas during the contraction process. What is V2? Let the ideal-gas constant R = 8.314 J/(mol .K).

Answer 1

V₂ = 14.07 L

Step-by-step explanation:

As this gas is cooled at constant temperature of 320 K, this means that we are on an isothermal process, and according to the 1st law of thermodynamics:

Q = W (1)

And as the temperature is constant, we can use the following expression to calculate the Work done:

W = nRT ln(V₁/V₂) (2)

However, as Q = W, we can replace heat into the above expression and then solve for V₂:

Q = nRT ln(V₁/V₂)

Replacing we have:

1200 = (1.5 * 8.314 * 320) ln(19/V₂)

1200 = 39907.2 ln(19/V₂)

ln(19/V₂) = 1200/3990.72

ln(19/V₂) = 0.3007

19/V₂ = e^(0.3007)

V₂ = 19 / e^(0.3007)

V₂ = 14.07 L

Hope this helps