The temperature of 3.00 , mol of an ideal dia | vedclass

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(A) Since the pressure of the gas remains constant,the process is isobaric.For a diatomic gas where molecules rotate but do not oscillate,the degrees

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) Since the pressure of the gas remains constant,the process is isobaric.For a diatomic gas where molecules rotate but do not oscillate,the degrees of freedom $f = 5$.The change in internal energy is given by $\Delta U = n C_v \Delta T = n \left( \frac{f}{2} R ight) \Delta T = \frac{5}{2} n R \Delta T$.The work done by the gas in an isobaric process is $W = P \Delta V = n R \Delta T$.The ratio of the change in internal energy to the work done is $\frac{\Delta U}{W} = \frac{\frac{5}{2} n R \Delta T}{n R \Delta T} = \frac{5}{2} = 2.5$.Given that this ratio is equal to $\frac{x}{10}$,we have $\frac{x}{10} = 2.5$.Therefore,$x = 2.5 	imes 10 = 25$.

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