ThermodynamicsHard

Question

One mole of an ideal monoatomic gas is heated in a process PV5/2 = constant. By what amount heat is absorbed in the process in 26°C rise in temperature?

Options

A.100 J
B.180 J
C.200 J
D.208 J

Solution

$C_{m} = C_{V,m} + \frac{R}{1 - x} = \frac{3R}{2} + \frac{R}{1 - \frac{5}{2}} = \frac{5R}{6}$

$\therefore q = n.C_{m}.\Delta T = 1 \times \frac{5R}{6} \times 26 = 180.14\text{ J}$

Create a free account to view solution

View Solution Free
Topic: Thermodynamics·Practice all Thermodynamics questions

More Thermodynamics Questions

Select the correct statement...ΔE must be zero for...The enthalpy and entropy change for a chemical reaction are −2500 cal and +7.4 cal/K, respectively. The nature of reacti...Any process will be spontaneous at constant pressure and temperature when...If a triatomic non-linear gas is heated isothermally, then what percentage of the heat energy is used to increase the in...