ThermodynamicsHard
Question
When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its temperature, the change in entropy is
Options
A.Cv,m ln2
B.Cp,m ln2
C.R ln2
D.(Cv,m − R) ln2
Solution
$\Delta S = n.C_{v,m}.\ln\frac{T_{2}}{T_{1}} + nR.\ln\frac{V_{2}}{V_{1}} = 1 \times C_{v,m} \times \ln 2 + 1 \times R \times \ln\frac{1}{2} = \left( C_{v,m} - R \right).\ln 2$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
Given the following entropy values (in J/K-mol) at 298 K and 1 atm H2(g) = 130.6, Cl2(g) = 223.0 and HCl(g) = 186.7. The...If four identical samples of an ideal gas initially at the same state (Po, Vo, To) are allowed to expand to double their...If one mole of a monoatomic gas ($\gamma$ = 5/3) is mixed with one mole of a diatomic gas ($\gamma$ = 7/5), then the val...If the enthalpy change for the transition of liquid water to steam is 30 kJ mol-1 at 27oC, the entropy change for the pr...Molar heat capacity of CD2O (deuterated form of formaldehyde) vapour at constant pressure is vapour 14 cal/K-mol. The en...