A 2 text kg sample of monoatomic gas is maintained at a pressure of 4 times 10^4 text N/m ^2 with a density of 8 text kg/m ^3. What is the order of the gas's internal energy due to thermal motion?

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EEJ MAIN Physics QUESTION #7057

Question 1

A $2 \text{ kg}$ sample of monoatomic gas is maintained at a pressure of $4 \times 10^4 \text{ N/m}^2$ with a density of $8 \text{ kg/m}^3$. What is the order of the gas's internal energy due to thermal motion?

$10^3 \text{ J}$
$10^5 \text{ J}$
$10^4 \text{ J}$✔️
$10^6 \text{ J}$

Correct Answer Explanation

The internal energy $U$ of a monoatomic gas is $U = \frac{3}{2} nRT = \frac{3}{2} PV$.

First, find the volume $V$ using mass and density: $V = \frac{\text{mass}}{\text{density}} = \frac{2}{8} = 0.25 \text{ m}^3$.

$U = \frac{3}{2} \times (4 \times 10^4) \times 0.25 = 1.5 \times 10^4 = 15000 \text{ J}$.

The order of magnitude is $10^4 \text{ J}$.

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