ThermodynamicsHard
Question
A liquid is adiabatically expanded from state – I to state – II, suddenly by a single step, as shown in the figure then
Options
A.$\Delta H = \frac{2\gamma P_{o}V_{o}}{\gamma - 1}$
B.$\Delta U = \frac{3P_{o}V_{o}}{\gamma - 1}$
C.ΔH = −PoVo
D.ΔU = −3PoVo
Solution
$q = 0 \Rightarrow \Delta U = w = - P_{0}\left( 4V_{0} - V_{0} \right) = - 3P_{0}V_{0}$
Now, $\Delta H = \Delta U + \Delta(PV) = \left( - 3P_{0}V_{0} \right) + \left( P_{0}.4V_{0} - 2P_{0}.V_{0} \right) = - P_{0}.V_{0}$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
The equation of state for one mole of a gas is PV = RT + BP, where B is a constant, independent of temperature. The inte...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...Calculate the equilibrium constant for the reaction :PCl5 ⇋ PCl3 + Cl2 (g) (g) (g)at 400K if ᐃHo = 69418.86 ...One mole of an ideal monoatomic gas is heated in a process PV5/2 = constant. By what amount heat is absorbed in the proc...Which of the following processes takes place with decrease of entropy ?...