ThermodynamicsHard
Question
A definite mass of a monoatomic ideal gas at 1 bar and 27°C expands against vacuum from 1.2 dm3 to 2.4 dm3. The change in free energy of the gas, ΔG, is (R = 0.08 bar- L/K-mol, ln 2 = 0.7)
Options
A.0
B.−64 bar-l
C.+84 J
D.−84 J
Solution
Free expansion is isothermal
$\Delta G = nRT\ln\frac{P_{2}}{P_{1}} = nRT\ln\frac{V_{2}}{V_{1}} $$$= 10^{5} \times \left( 1.2 \times 10^{- 3} \right) \times \ln\frac{1.2}{2.4} = - 84\text{ J}$$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
In an irreversible process taking place at constant T and P and in which only pressure-volume work is being done, the ch...Standard enthalpy of vaporisation ᐃvapHo for water at 100oC is 40.66 kJ mol-1. The internal energy of vaporisation...An ideal monoatomic gas undergoes a reversible process, where $\frac{P}{V}$ = constant, from (2 bar, 273 K) to 4 bar. Th...For a system in equilibrium, ΔG = 0 under conditions of constant...A heat engine absorbs heat Q1 at temperature T1 and heat Q2 at temperature T2. Work done by the engine is (Q1 + Q2). Thi...