Chemistry - Thermodynamics - I Question with Solution | TestHub

Explain the question: Find the change in enthalpy ($\Delta \mathrm{H}$) during the reversible adiabatic expansion stage of the given three-step thermodynamic cycle.

Concept: Reversible adiabatic process, ${\mathrm{T}}_1{\mathrm{V}}_1^{\gamma -1}={\mathrm{T}}_2{\mathrm{V}}_2^{\gamma -1}$.

Solution:

The gas starts this step at T₂ = 600 K and V₂ = 2V₁ and expands to a final volume V₃ =

$$4\sqrt{2}{V}_1$$

Using the temperature-volume relationship (TVᵞ⁻¹ = const with $\gamma -1=\frac{2}{3}$):

$600\cdot (2V_1{)}^{2/3}=T_3\cdot {\left(4\sqrt{2}{V}_1\right)}^{\frac{2}{3}}$

$T_3=600\times {\left(\frac{2V_1}{4\sqrt{2}{V}_1}\right)}^{\frac{2}{3}}=600\times {\left(\frac{1}{2\sqrt{2}}\right)}^{\frac{2}{3}}=600\times {\left(\frac{1}{2^{3/2}}\right)}^{\frac{2}{3}}$

$T_3=600\times {\left(2^{-3/2}\right)}^{\frac{2}{3}}=600\times 2^{-1}=300K$

For an ideal monatomic gas,

$\Delta H=2.5R\times (300K-600K)=2.5R\times (-300)=-750R$