Question

One litre of hydrogen at 27°C and 10⁶ dyne cm^-2 pressure expands isothermally until its volume in doubled and then adiabatically until redoubled. Find the final temperature, pressure and work done in each case, γ = 1.4.

Answer 1

Given :

P₁ = 10⁶ dyne cm^-2

V₁ = 103 cm³ = 1 litre

T₁ = 27°C = 300 K

V₂ = 2V₁ = 2000 cm³ for isothermal expansion

T₂ = ?

For adiabatic expansion P₂ = ?

V₁' =  2000 cm³

V₂' = 2₁' = 4000 cm³

(a) For isothermal expansion :

P₂V₂ = P₁V₁

or P₂ = \(\frac{P_1V_1}{V_2}\)

= \(\frac{10^6\times10^3}{2\times10^3}\)

= 5 × 10⁵ dyne cm^-2

T₂ = T₁ for isothermal expansion = 300 K = 27°C Work done during isothermal expansion is given by

W_iso = 2.303 RT log₁₀ \((\frac{V_2}{V_1})\)

or  W_iso = 2.303 P₁V₁ log₁₀\((\frac{V_2}{V_1})\) \((\frac{V_2}{V_1})\) 

or  W_iso = 2.303 × 10⁶ × 10³log₁₀

or  W_iso = 2.303 × 10⁹ × 0.3010

or  W_iso = 6.93 × 10⁸ erg.

(b) For adiabatic expansion:

P₂ = P₁\((\frac{V_1'}{V_2'})^γ\)

= 10⁶\((\frac{2000}{4000})^{1.4}\)

or P₂ = 10⁶ ×\((\frac{1}{2})^{1.4}\) 

or P₂ = 1.895 × 10⁵ dyne cm^-2

From relation T₁V₁^γ−1 = ₂₂^γ−1, we get 

T₂ = T₁\((\frac{V_1}{V_2})^{γ-1}\) 

= 300 \((\frac{1}{2})^{1.4}\) 

= 227.4K = −45.6°C

Work done during adiabatic expansion is given by

W_adi = (T₁ − T₂ )\(\frac{R}{1-γ}\)

\(= \frac{P_1V_1-P_2V_2}{γ-1}\)

\(= \frac{10^6\times2\times10^3-1.895\times10^5\times4\times10^3}{1.4-1}\)

or W_adi = \( \frac{20\times10^8-1.895\times4\times10^8}{0.4}\)

or W_adi =  31.05 × 10⁸ erg.