If kinetic energy of one mole of monoatomic gas at $27^{\circ} C$ is $E$ joules, what will be the internal energy of one mole of diatomic gas at $327^{\circ} C$ ?

\frac{5 E}{3} J

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\frac{10 E}{3} J

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\frac{3 E}{4} J

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\frac{5 E}{4} J

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Solution

The correct option is B $\frac{10 E}{3} J$
$U = n C_{V} T (C_{V} = \frac{f R}{2}) (f = d e g r e e o f f r e e d o m) E_{m o n o} = \frac{3}{2} n R T$ $(f = 3)$
$U = E_{d i a t o m i c} = \frac{5}{2} n R T^{'}$ $(f = 5)$
$\frac{U}{E} = \frac{\frac{5}{2} n R T^{'}}{\frac{3}{2} n R T} = \frac{5 \times 600}{3 \times 300} = \frac{30}{9} = \frac{10}{3}$
$U = \frac{10 E}{3}$

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If kinetic energy of one mole of monoatomic gas at 27 ∘C is E joules, what will be the internal energy of one mole of diatomic gas at 327 ∘C ?

The correct option is B 10E3 J U=nCVT ( CV=fR2 )(f=degree of freedom)Emono=32nRT (f=3) U=Ediatomic=52nRT′ (f=5) UE=52 nRT′32 nRT=5×6003×300=309=103 U=10E3

If kinetic energy of one mole of monoatomic gas at $27^{\circ} C$ is $E$ joules, what will be the internal energy of one mole of diatomic gas at $327^{\circ} C$ ?