The total internal energy of 2 moles of a mon | vedclass

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(B) For a monatomic gas,the degrees of freedom $f_1 = 3$. Given $n_1 = 2$ moles and $T_1 = 27^{\circ} C = 300 \ K$. The internal energy $U$ is given b

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$\frac{10 U}{3}$

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(B) For a monatomic gas,the degrees of freedom $f_1 = 3$. Given $n_1 = 2$ moles and $T_1 = 27^{\circ} C = 300 \ K$. The internal energy $U$ is given by $U = \frac{n_1 f_1 R T_1}{2} = \frac{2 	imes 3 	imes R 	imes 300}{2} = 900 R$.For a diatomic gas,the degrees of freedom $f_2 = 5$. Given $n_2 = 3$ moles and $T_2 = 127^{\circ} C = 400 \ K$. The internal energy $U'$ is given by $U' = \frac{n_2 f_2 R T_2}{2} = \frac{3 	imes 5 	imes R 	imes 400}{2} = 3000 R$.Now,calculating the ratio: $\frac{U'}{U} = \frac{3000 R}{900 R} = \frac{30}{9} = \frac{10}{3}$.Therefore,$U' = \frac{10 U}{3}$.

The total internal energy of $2$ moles of a monatomic gas at a temperature $27^{\circ} C$ is $U$. The total internal energy of $3$ moles of a diatomic gas at a temperature $127^{\circ} C$ is:

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