Internal energy of n₁ moles of hydrogen at temperature ' T ' is equal to internal energy of ' n₂ ' moles of helium at temperature 2T, then the ratio n₁: n₂ is [Degree of freedom of He = 3, Degree of freedom of H₂ = 5 ] (a) 5: 3 (b) 6: 5 (c) 2: 3 (d) 3: 5

Answer: (b). U = nf/2RT. n₁ · 5/2R T = n₂ · 3/2R(2T) 5n₁ = 6n₂ ⇒ n₁: n₂ = 6: 5 So the answer is (b) 6: 5.

Internal energy of n₁ moles of hydrogen at temperature ' T ' is equal… | MHT CET 2021 (24 Sep, Shift 1)

Q23MHT CET 2021 · 24 SEP · SHIFT 1Medium

Degrees of Freedom & Internal Energy

Internal energy of $n_1$ moles of hydrogen at temperature ' $T$ ' is equal to internal energy of ' $n_2$ ' moles of helium at temperature $2T$ , then the ratio $n_1 : n_2$ is

[Degree of freedom of $\text{He} = 3$ , Degree of freedom of $\text{H}_2 = 5$ ]

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Answer: (b)

$U = n\frac{f}{2}RT$ .
$n_1 \cdot \frac{5}{2}R T = n_2 \cdot \frac{3}{2}R(2T)$
$5n_1 = 6n_2 \Rightarrow n_1 : n_2 = 6 : 5$
So the answer is (b) $6 : 5$ .

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Degrees of Freedom & Internal Energy question from MHT CET 2021 (24 Sep, Shift 1)

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