Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IIGibb's Free EnergyEasy2 minQB
ChemistryEasynumerical
The increase in Gibbs free energy (in kJ) of 32 g of methanol (density = 0.80 g cm⁻³), when the pressure is increased isothermally from 1 bar to 2001 bar, is
Answer:
8.00
Solution:
ΔG = VΔP. V = 32/0.80 = 40 cm³ = 40×10⁻⁶ m³; ΔP = 2000 bar = 2×10⁸ Pa. ΔG = 40×10⁻⁶ × 2×10⁸ = 8000 J = 8 kJ.
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Gibb's Free Energy
⏱ 2mℹ️ Source: QB
Doubts & Discussion
Loading discussions...
