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The volume (V) of a monatomic 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

Option 1)
$\frac{1}{3}$

Option 2)
$\frac{2}{3}$

Option 3)
$\frac{2}{5}$

Option 4)
$\frac{2}{7}$

Answers (2)

best_answer

As we have learned

Bulk Modulus of elasticity in isobaric process -

$omega = int p: dV$

$=pleft ( V_{f}-V_{i} ight )$

- wherein

$W =pleft ( V_{f}-V_{i} ight )$

$=nRleft ( T_{f}-T_{i} ight )$

First law in isobaric process -

$Delta U= n, C_{v}Delta T$

$= nfrac{R}{gamma -1}Delta T$

- wherein

$Delta Q= Delta U+W$

$= nfrac{gamma : R}{gamma -1}cdot Delta T$

$Delta Q= nC_{p}: Delta T$

$V\propto T$

$V= KT$

$PV= nRT$

or $p=\frac{nRT}{KT}=\frac{nR}{K}$

work done =

$= P\frac{nR}{K}(\Delta V)= \frac{nR}{K}(K\Delta T)= n R\Delta T$

heat absorbed = $nC_{p}\Delta T$

$=n(\frac{5R}{2})\Delta T= \frac{5}{2}(nR\Delta T)$

ratio of work to heat

$\frac{nR\Delta T}{5/2 nR\Delta T}=2/5$

Option 1)

$\frac{1}{3}$

This is incorrect

Option 2)

$\frac{2}{3}$

This is incorrect

Option 3)

$\frac{2}{5}$

This is correct

Option 4)

$\frac{2}{7}$

This is incorrect

Posted by

Plabita

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OPTION 3

Posted by

Md Shams tabrez

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