In the reversible adiabatic expansion of a gas the increase in disorder due to an increase in volume is compensated by the decrease in disorder due to a decrease in temperature.

In the reversible adiabatic expansion of a gas the increase in disorder due to an increase in volume is compensated by the decrease in disorder due to a decrease in temperature. Correct Answer true

This ensures that the disorder number or entropy remains constant.

Related Questions

On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at a point A, the slope of the reversible adiabatic line $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{S}}}$$  and the slope of the reversible isothermal line $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{T}}}$$  are related as (where, $${\text{y}} = \frac{{{{\text{C}}_{\text{p}}}}}{{{{\text{C}}_{\text{v}}}}}$$  )
Work-done during adiabatic expansion is given by (where p1, v1, T1 = Pressure, volume and temperature for the initial condition of gas, p2, v2, T2 = Corresponding values for the final condition of gas, R = Gas constant and $$\gamma $$ = Ratio of specific heats)