First slide

Different type of processes

Question

The volume (V) of a monoatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

Moderate

A

$\frac{2}{5}$
B

$\frac{2}{3}$
C

$\frac{2}{7}$
D

$\frac{2}{9}$

Solution

$V = \left( {\frac{{nR}}{P}} \right)T$
For the diagram, $V \propto T$
So we conclude that P = constant.
Hence the process AB is isobaric.
Now, ΔQ = ΔU + ΔW
$\therefore \frac{{\Delta W}}{{\Delta Q}} = \frac{{\Delta Q - \Delta U}}{{\Delta Q}} = 1 - \frac{{\Delta U}}{{\Delta Q}} = 1 - \frac{{{C_V}}}{{{C_P}}}$
$= 1 - \frac{1}{\gamma }$
$\gamma = 1 + \frac{2}{3} = \frac{5}{3}$
$\therefore \frac{{\Delta W}}{{\Delta Q}} = 1 - \frac{3}{5} = \frac{2}{5}.$

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