A polyatomic gas with 6 degrees of freedom do | vedclass

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(A) Given: Degrees of freedom $f = 6$,Work done $\Delta W = 25 \ J$.For an ideal gas,the adiabatic index $\gamma$ is given by $\gamma = 1 + \frac{2}{f

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$100$

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(A) Given: Degrees of freedom $f = 6$,Work done $\Delta W = 25 \ J$.For an ideal gas,the adiabatic index $\gamma$ is given by $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3}$.At constant pressure,the work done is $\Delta W = P \Delta V = nR \Delta T$.The heat supplied is $\Delta Q = n C_p \Delta T$.We know that $C_p = \frac{\gamma R}{\gamma - 1}$.Thus,$\frac{\Delta W}{\Delta Q} = \frac{nR \Delta T}{n C_p \Delta T} = \frac{R}{C_p} = \frac{R}{\frac{\gamma R}{\gamma - 1}} = \frac{\gamma - 1}{\gamma}$.Substituting $\gamma = \frac{4}{3}$:$\frac{\Delta W}{\Delta Q} = \frac{\frac{4}{3} - 1}{\frac{4}{3}} = \frac{1/3}{4/3} = \frac{1}{4}$.Therefore,$\Delta Q = 4 	imes \Delta W = 4 	imes 25 \ J = 100 \ J$.

$A$ polyatomic gas with $6$ degrees of freedom does $25 \ J$ of work when it is expanded at constant pressure. The heat given to the gas is ..... $J$.

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