Gujarati
Hindi
11.Thermodynamics
normal

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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