ThermodynamicsHard

Question

Two moles of an ideal gas ($\gamma$ = 1.4) was allowed to expand reversibly and adiabatically from 1 L, 527°C to 32 L. The molar enthalpy change of the gas is

Options

A.−4200R
B.−2100R
C.−1500R
D.−3000R

Solution

$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1} \Rightarrow T_{2} = T_{1}\left( \frac{V_{1}}{V_{2}} \right)^{\gamma - 1} = 800 \times \left( \frac{1}{32} \right)^{\frac{7}{5} - 1} = 200\text{ K}$

$\therefore\Delta H = n.C_{p,m}.\left( T_{2} - T_{1} \right) = 1 \times \frac{7}{2}R \times (200 - 800) = - 2100\text{ R}$

Create a free account to view solution

View Solution Free
Topic: Thermodynamics·Practice all Thermodynamics questions

More Thermodynamics Questions