ThermodynamicsHard
Question
The entropy change of 2.0 moles of an ideal gas whose adiabatic exponent $\gamma$ = 1.50, if as a result of a certain process, the gas volume increased 2.0 times while the pressure dropped 4.0 times, is (ln 2 = 0.7)
Options
A.−11.64 J/K
B.+11.64 J/K
C.−34.92 J/K
D.+34.92 J/K
Solution
$\Delta S = n.C_{V,m}.\ln\frac{T_{2}}{T_{1}} + nR.\ln\frac{V_{2}}{V_{1}}$
$= n.C_{V,m}.\ln\frac{P_{2}}{P_{1}} + n.C_{P,m}.\ln\frac{V_{2}}{V_{1}} $$$= 2 \times \frac{R}{1.5 - 1} \times \ln\frac{1}{4} + 2 \times \frac{1.5R}{1.5 - 1} \times \ln 2 = - 11.64\text{ J/K}$$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
Two moles of an ideal monoatomic gas is heated from 27°C to 627°C, reversibly and isochorically. The entropy of gas...Molar heat capacity of CD2O (deuterated form of formaldehyde) vapour at constant pressure is vapour 14 cal/K-mol. The en...For a process to occur under isothermal conditions, the essential condition(s) is/ are...A student is calculating the work done by 2 mole of an ideal gas in a reversible isothermal expansion shown in the figur...At 500 kbar and T K, the densities of graphite and diamond are 2.0 and 3.0 g/cm3, respectively. The value of (ΔH − ΔU) f...