ThermodynamicsHard
Question
A quantity of 1.6 g helium gas is expanded adiabatically 3.0 times and then compressed isobarically to the initial volume. Assume ideal behaviour of gas and both the processes to be reversible. The entropy change of the gas in this process is (ln 3 = 1.1)
Options
A.−1.1 cal/K
B.+1.1 cal/K
C.−2.2 cal/K
D.+2.2 cal/K
Solution
$\Delta S = \Delta S_{\text{adiabatic}} + \Delta S_{isobaric} = 0 + n.C_{P,m}.\ln\frac{T_{2}}{T_{1}}$
$= \frac{1.6}{4} \times \frac{5R}{2} \times \ln\frac{1}{3} = - 2.2\text{ Cal/K}$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
One mole of a non-ideal gas undergoes a change of state (2 atm, 3 L, 95 K) → (4 atm, 5 L, 245 K) with a change in intern...A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37....The efficiency of a Carnot cycle is 1/6. On decreasing the temperature of the sink by 65°C, the efficiency increases to ...The change that does not increase entropy is...The internal energy of an ideal gas increases during an isothermal process when the gas is...