Question

Suppose 1400 J of heat are added to 1.8 mol of argon gas at a constant pressure of 120 kPa. (Assume that the argon can be treated as an ideal monatomic gas.)

(a) Find the change in internal energy. J

(b) Find the change in temperature for this gas. K

(c) Calculate the change in volume of the gas. m3

Answer 1

Solution:

Given:

Q = 1400 J

n = 1.8 mol

P = 120 kPa

Since: Q = (5/2) n R ΔT

1400 = (5/2)(1.8)(8.314) ΔT

ΔT = 37.42 K

The change in internal energy is : ΔU = (3/2) n R ΔT

ΔU = (3/2)(1.8 mol)(8.314)(37.42)

ΔU = 840 J

Since: Q = ΔU + W

Thus : The work done is given as : W = 1400 - 840 = 560 J

But, W = P ΔV

560 = (120 x 10³ Pa) ΔV

ΔV = 0.004666 m³