Question

5 moles of helium expand isothermally and reversibly from a pressure 40 × 10^-5  N m^-2 to 4 × 10^-5 N m^-2  at 300 K. Calculate the work done, change in internal energy and heat absorbed during the expansion. (R = 8.314 J K^-1  mol^-1).

Answer 1

Given:
Number of moles of helium gas (n) = 5
Initial pressure (P₁) = 40 x 10^-5 Nm^-2
Final pressure (P₂) = 4 x 10^-5 Nm^-2
Temperature (T) = 300 K
R = 8.314 J K^-1 mol^-1

To find:
a. Work done (W)
b. Change in internal energy (ΔU)
c. Heat absorbed ( q)

Formulae: 
a. W_max = - 2.303 nRT log10 `"P"_1/"P"_2`

b. ΔU = q + W

Calculation:
a. From formula (a),
 W_max = - 2.303 nRT log10 `"P"_1/"P"_2`

= - 2.303 x 5 x 8.314 x 300 `log_10 ( 40 xx 10^-5)/(4 xx 10^-5)`

= -2.303 x 5 x 8.314 x 300 log₁₀ 10

= - 28720.71 J 
= - 28.72 kJ
∴ Work done (W) = - 28.72 KJ

b. Form formula (b),
According to first law of thermodynamics, ΔU = q + W
In isothermal process, ΔT = 0, hence ΔU = 0.
∴ Change in internal energy (ΔU) = 0

c. From formula (b),
0 = q + W
∴ q = - W
= - ( - 28.72 kJ )
= 28.72 kJ
∴ Heat absorbed (q) = 28.72 kJ.