Question

A 2.20-mol sample of helium gas initially at 300 k, and 0.400 atm is compressed isothermally to 1.80 atm. Note that the helium behaves as an ideal gas. (a) find the final volume of the gas. M3 (b) find the work done on the gas. Kj (c) find the energy transferred by heat. Kj

Answer 1

use ideal gas law

p·V = n·R·T

=>

V = n·R·T/p

= 2.2mol · 0.08205784 Latm/molK · 300K / 1.5atm

= 36.3L

(b)

work done on gas is given by the work integral:

W = - ∫ (V1→V2) p dV

= - ∫ (V1→V2) n·R·T/V dV

= - n·R·T·ln(V2/V1) (for T= constant)

= - n·R·T·ln(p1/p2)

=>

W = - 2.2mol · 8,314472J/molK · 300k · ln(0.4atm / 1.5atm)

= 7253.2J

(c)

The change of internal energy is equal to to the sum of work done and heat transferred to the gas:

Δ U = W + Q

The internal energy remains unchanged for isothermal processes on ideal gases:

ΔU = ∫ n·Cv dT = 0

<=>

W + Q = 0

<=>

Q = - W = -7253.2J

Negative sign indicates that this amount of heat is removed from the gas.

Answer 2

Answer:

Explanation:

Refer the below attachment.