Question

find the magnitude and ratio of internal energies for 5moles of mono-atomic gas to the di-atomic gas at temperature 27c

Answer 1

Answer on Question #43120 – Physics – Molecular Physics | Thermodynamics

Question.

Find the magnitude and ratio of internal energies for 5 moles of mono-atomic gas to the di-atomic gas at temperature 27°C.

Given:

$\nu = 5$ moles

$T = 27{}^{\circ}C = 300\ K$

Find:

$U_{1} = ?\ U_{2} = ?$

$\frac{U_{2}}{U_{1}} = ?$

Solution.

From the thermodynamics it’s known that internal energy is:

$$U = \frac{i}{2} \nu R T$$

$i$ is the number of degrees of freedom;

$R = 8.31 \frac{J}{mole \cdot K}$ is the gas constant.

For monoatomic gas $i = 3$, for diatomic gas $i = 5$.

So,

$$U_{1} = \frac{3}{2} \nu R T;\ U_{2} = \frac{5}{2} \nu R T$$

Calculate:

$$U_{1} = \frac{3}{2} 5 \cdot 8.31 \cdot 300 = \frac{3}{2} 12465 = 18697.5\ J$$ $$U_{2} = \frac{5}{2} 5 \cdot 8.31 \cdot 300 = \frac{5}{2} 12465 = 31162.5\ J$$

And

$$\frac{U_{2}}{U_{1}} = \frac{5}{3}$$

Answer.

$$U _ {1} = \frac {3}{2} \nu R T = 1 8 6 9 7. 5 J$$ $$U _ {2} = \frac {5}{2} \nu R T = 3 1 1 6 2. 5 J$$ $$\frac {U _ {2}}{U _ {1}} = \frac {5}{3}$$

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