ThermodynamicsHard
Question
Consider a reversible isentropic expansion of 1 mole of an ideal monoatomic gas from 27°C to 927°C. If the initial pressure of gas was 1 bar, then the final pressure of gas becomes
Options
A.4 bar
B.8 bar
C.32 bar
D.0.25 bar
Solution
$\Delta S = n.C_{P,m}.\ln\frac{T_{2}}{T_{1}} + nR.\ln\frac{P_{1}}{P_{2}}$
$\text{Or, }0 = n \times \frac{5}{2}R \times \ln\frac{1200}{300} + nR \times \ln\frac{1}{P_{2}} \Rightarrow P_{2} = 32\text{ bar}$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
For an ideal gas subjected to different processes as shown in the graphs, select the graph which will involve the greate...Choose the substance which has higher possible entropy (per mole) at a given temperature....Two moles of an ideal monoatomic gas is heated from 27°C to 627°C, reversibly and isochorically. The entropy of gas...An ideal gas with the adiabatic exponent $\gamma$goes through a process P = Po – αV, where Po and α are positive constan...For which of the following ideal gas, Cv,m is independent of temperature?...