First slide

First law of thermodynamics

Question

When a quantity of heat ‘Q’ is supplied to a monoatomic gas work done by the gas is $\frac{Q}{4}$ Molar heat capacity of the gas in that process is

Moderate

A

$\frac{10 R}{3}$
B

$\frac{5 R}{2}$
C

2 R
D

$\frac{1}{2} R$

Solution

if $\frac{Q}{4}$ heat is the work done then
$\frac{3 Q}{4}$ is the heat for internal energy
$\frac{3 Q}{4} = {nC}_{v} dT$
$\frac{3 Q}{4} = n \frac{3 R}{2} dT$ (monoatomic gas)
$(\frac{Q}{ndT}) \to$ molar specific heat - $\frac{Q}{ndT} = 2 R$

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