Question 903: Thermodynamics & KTG

Question

Internal energy of $$n_1$$ moles of hydrogen at temperature '$$T$$' is equal to internal energy of '$$n_2$$' moles of helium at temperature $$2 T$$, then the ratio $$\mathrm{n}_1: \mathrm{n}_2$$ is
[Degree of freedom of $$\mathrm{He}=3$$, Degree of freedom of $$\mathrm{H}_2=5$$]

Accepted Answer

The internal energy of an ideal gas is given by the following formula:
$$U = \frac{f}{2} nRT$$
where:
$$U$$ is the internal energy
$$f$$ is the degree of freedom
$$n$$ is the number of moles
$$R$$ is the ideal gas constant
$$T$$ is the temperature
We are given that the internal energy of $$n_1$$ moles of hydrogen at temperature $$T$$ is equal to the internal energy of $$n_2$$ moles of helium at temperature $$2T$$.  Therefore, we can write the following equation:
$$\frac{5}{2}n_1RT = \frac{3}{2}n_2R(2T)$$
Simplifying the equation, we get:
$$5n_1 = 6n_2$$
Therefore, the ratio $$n_1: n_2$$ is $$\boxed{6:5}$$.

Answer

$$5: 3$$

Answer

$$6: 5$$

Answer

$$2: 3$$

Answer

$$3: 5$$

Thermodynamics & KTG

Internal energy of $n_1 moles of hydrogen at temperature 'T' is equal to internal energy of 'n_2' moles of helium at temperature 2 T, then the ratio \mathrm{n}_1: \mathrm{n}_2 is [Degree of freedom of \mathrm{He}=3, Degree of freedom of \mathrm{H}_2=5$]

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