ThermodynamicsHard
Question
A student is calculating the work done by 2 mole of an ideal gas in a reversible isothermal expansion shown in the figure. By mistake he calculated the area of the shaded area in the PV graph shown, as work and answered the magnitude of work equal to 49.26 L-atm. What is the correct magnitude of work done by the gas in L-atm? (R = 0.0821 L-atm/K-mol)
Options
A.49.26
B.98.52
C.78.63
D.34.14
Solution
Area = P2 × ΔV ⇒ P2 × 4 = 49.26 L-atom
Now, correct work, $w = - nRT.\ln\frac{V_{2}}{V_{1}} = - P_{2}V_{2}.\ln\frac{4}{2}$= –49.26 × 0.693 = – 34.137 L-atom
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
When the value of entropy is greater, then the ability for work is...Which one of the following has ᐃSogreater than zero ?...Among the following, the intensive property is (properties are)...The plot of $\log_{10}\text{ }K$ vs $\frac{1}{\text{ }T}$ gives a straight line. The intercept and slope respectively ar...One mole of monoatomic ideal gas at T K is expanded from 1 L to 2 L adiabatically under a constant external pressure of ...