A diatomic gas (gamma=1.4) does 100 ,J of wor | vedclass

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Question

(A) For an isobaric process, the work done is given by $W = P \Delta V = nR \Delta T = 100 \,J$.The heat supplied to the gas is given by the first law

Vedclass · Accepted Answer

$350$

Vedclass · Answer

(A) For an isobaric process, the work done is given by $W = P \Delta V = nR \Delta T = 100 \,J$.The heat supplied to the gas is given by the first law of thermodynamics: $Q = \Delta U + W$.For a diatomic gas, the degrees of freedom $f = 5$. The change in internal energy is $\Delta U = \frac{f}{2} nR \Delta T = \frac{5}{2} nR \Delta T$.Substituting $\Delta U$ and $W$ into the heat equation: $Q = \frac{5}{2} nR \Delta T + nR \Delta T = \left(\frac{5}{2} + 1ight) nR \Delta T = \frac{7}{2} nR \Delta T$.Since $nR \Delta T = 100 \,J$, we have $Q = \frac{7}{2} 	imes 100 = 350 \,J$.

$A$ diatomic gas $(\gamma=1.4)$ does $100 \,J$ of work in an isobaric expansion. The heat given to the gas is: (in $\,J$)

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