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
$C_{m} = C_{V,m} + \frac{R}{1 - x} = \frac{3R}{2} + \frac{R}{1 - \frac{5}{2}} = \frac{5R}{6}$
$\therefore q = n.C_{m}.\Delta T = 1 \times \frac{5R}{6} \times 26 = 180.14\text{ J}$
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