Question 864: Thermodynamics & KTG

Question

An ideal gas having molar mass '$$\mathrm{M}_0$$', has r.m.s. velocity 'V' at temperature 'T'. Then

Accepted Answer

To solve this question, we need to recall the relationship between the root mean square (r.m.s.) velocity of an ideal gas, its temperature, and its molar mass. The r.m.s. velocity, $$V$$, of an ideal gas can be expressed using the equation:
$$V = \sqrt{\frac{3RT}{M_0}}$$
where:
$$V$$ is the root mean square velocity,
$$R$$ is the universal gas constant,
$$T$$ is the temperature in Kelvin,
$$M_0$$ is the molar mass of the gas.
We must analyze each option in light of this equation.
Option A: $$\mathrm{VT}^2=$$ constant
We substitute the expression for $$V$$ into this option:
$$\mathrm{VT}^2 = \left(\sqrt{\frac{3RT}{M_0}}\right)T^2 = \sqrt{\frac{3R}{M_0}}T^{2.5}$$
Here, we see that $$\mathrm{VT}^2$$ is not a constant because it is proportional to $$T^{2.5}$$.
Option B: $$\frac{\mathrm{v}^2}{\mathrm{T}}=$$ constant
Notice that there's a typographic mistake in the question: it should probably refer to $$V^2$$ instead of $$\mathrm{v}^2$$ for consistency. Correcting this and substituting the expression for $$V$$, we get:
$$\frac{V^2}{T} = \frac{\frac{3RT}{M_0}}{T} = \frac{3R}{M_0}$$
This expression clearly shows that $$\frac{V^2}{T}$$ is a constant, because it only depends on constants $$R$$ and $$M_0$$, and does not change with temperature $$T$$.
Option C: $$\mathrm{V}^2 \mathrm{T}=$$ constant
Substituting the expression for $$V$$ gives:
$$V^2T = \left(\frac{3RT}{M_0}\right)T = \frac{3R}{M_0}T^2$$
Since this is proportional to $$T^2$$, it is not a constant relationship with respect to $$T$$.
Option D: $$\mathrm{V}$$ is independent of $$\mathrm{T}$$
Given the formula $$V = \sqrt{\frac{3RT}{M_0}}$$, we see that $$V$$ depends on $$T$$ (temperature), so this option is incorrect.
Therefore,
Option B
is correct: $$\frac{V^2}{T}$$ is a constant.

Answer

$$\mathrm{VT}^2=$$ constant

Answer

$$\frac{\mathrm{v}^2}{\mathrm{~T}}=$$ constant

Answer

$$\mathrm{V}^2 \mathrm{T}=$$ constant

Answer

$$\mathrm{V}$$ is independent of $$\mathrm{T}$$

Thermodynamics & KTG

An ideal gas having molar mass '$\mathrm{M}_0$', has r.m.s. velocity 'V' at temperature 'T'. Then

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