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11.Thermodynamics
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The average degree of freedom per molecule of a gas is $6$ . The gas performs $25\ J$ work, while expanding at constant pressure. The heat absorbed by the gas is .... $J$
A
$75$
B
$100$
C
$150$
D
$125$
Solution
For constant pressure process.
Work done $(\mathrm{W})=\mathrm{nR} \Delta \mathrm{T}$
$\therefore \Delta \mathrm{T}=\left(\frac{25}{\mathrm{nR}}\right)$
Now, for above process $f=6$
So $C_{P}\left(1+\frac{f}{2}\right) R=4 R$
$\text { Heat absorbed }=\mathrm{nC}_{\mathrm{P}} \Delta \mathrm{T}$
$=\mathrm{n} \times 4 \mathrm{R} \times \frac{25}{\mathrm{n} \mathrm{R}}=100 \mathrm{J}$
Standard 11
Physics
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