Question 280: Thermodynamics & KTG

Question

The temperature of an ideal gas is increased from $$200 \mathrm{~K}$$ to $$800 \mathrm{~K}$$. If r.m.s. speed of gas at $$200 \mathrm{~K}$$ is $$v_{0}$$. Then, r.m.s. speed of the gas at $$800 \mathrm{~K}$$ will be:

Accepted Answer

The root-mean-square (r.m.s) speed of an ideal gas is given by the formula:
$$v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}$$
where R is the gas constant, T is the temperature in Kelvin, and M is the molar mass of the gas.
In this case, we are given that the initial temperature is $$200 \, K$$ and the final temperature is $$800 \, K$$. Let the r.m.s speed at $$200 \, K$$ be $$v_0$$, then:
$$v_0 = \sqrt{\frac{3R \cdot 200}{M}}$$
Now, we want to find the r.m.s speed at $$800 \, K$$, let's call this $$v_1$$:
$$v_1 = \sqrt{\frac{3R \cdot 800}{M}}$$
Now, divide $$v_1$$ by $$v_0$$:
$$\frac{v_1}{v_0} = \frac{\sqrt{\frac{3R \cdot 800}{M}}}{\sqrt{\frac{3R \cdot 200}{M}}}$$
Simplify the expression:
$$\frac{v_1}{v_0} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2$$
So, $$v_1 = 2v_0$$.
Hence, the r.m.s speed of the gas at $$800 \, K$$ will be 2 times the r.m.s speed of the gas at $$200 \, K$$.

Answer

$$v_{0}$$

Answer

$$2 v_{0}$$

Answer

$$4 v_{0}$$

Answer

$$\frac{v_{0}}{4}$$

Thermodynamics & KTG

The temperature of an ideal gas is increased from $200 \mathrm{~K} to 800 \mathrm{~K}. If r.m.s. speed of gas at 200 \mathrm{~K} is v_{0}. Then, r.m.s. speed of the gas at 800 \mathrm{~K}$ will be:

Choose the correct answer: