Question

(a) What is the free expansion? Determine work done in case of free expansion of an ideal gas. 

(b) 4.0 mol of ideal gas at 2 atm and 25°C expands isothermally to 2 times of its original volume against the external pressure of 1 atm. Calculate work done. If the same gas expands isothermally in a reversible manner, then what will be the value of work done?

Answer 1

(a) When a gas expands under vacuum i.e. no external pressure work on it (P_ex = 0), its expansion is called free expansion.

\(\because\) P_ex = 0

\(\therefore\) W = 0 in case of free expansion of an ideal gas [\(\because\) W= -P_ex (V₂ - V₁) = 0]

It means no work is done

(b) Given

n = 4 moles

P = 2 atm

T = 25 + 273 = 298 K

P_ex = 1 atm

\(\therefore\) W = -P_ex(V_f - V_i)

Now, initial (V_i) =\(\frac{nRT}{P}\)

= \(\frac{4\times0.082\times298}{2}\)

= 48.87 L

\(\because\) Volume becomes 2 times of its original volume.

\(\therefore\) Final volume (V_f = 48.87 × 2 = 97.74 L)

\(\therefore\) W = −1(97.74 − 48.87)

= −1(48.87)

= −48.87 J

For isothermal reversible expansion of ideal gas

W_rev = -2.303 nRT log\(\frac{V_f}{V_i}\)

= −2.303 × 4 × 8.314 × 298 log\(\big(\frac{97.74}{48.87}\big)\)

= −2.303 × 4 × 8.314 × 298 × 0.3010

= −6869.84 J