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(B) Given,the degree of freedom $f = 6$.The adiabatic index $\gamma$ is given by $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac

Vedclass · Accepted Answer

$100$

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(B) Given,the degree of freedom $f = 6$.The adiabatic index $\gamma$ is given by $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3}$.For an isobaric process (constant pressure),the work done is $\Delta W = P \Delta V = nR \Delta T$.The heat absorbed is $\Delta Q = n C_p \Delta T = n \left( \frac{f+2}{2} R ight) \Delta T$.We know that $C_p = \frac{\gamma R}{\gamma - 1}$ and $R = C_p - C_v$,so $\frac{\Delta W}{\Delta Q} = \frac{nR \Delta T}{n C_p \Delta T} = \frac{R}{C_p} = \frac{C_p - C_v}{C_p} = 1 - \frac{1}{\gamma}$.Substituting the values: $\frac{25}{\Delta Q} = 1 - \frac{1}{4/3} = 1 - \frac{3}{4} = \frac{1}{4}$.Therefore,$\Delta Q = 25 	imes 4 = 100 \ J$.

The average degree of freedom per molecule of a gas is $6$. The gas performs $25 \ J$ of work while expanding at constant pressure. The heat absorbed by the gas is .... $J$.

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