Question

The root mean square speed of a gas molecule at $20{}^{\circ }\text{C}$ is $v$. If the temperature of the gas is raised by $10{}^{\circ }\text{C}$, then the root mean square speed of the gas molecule will be

• A. $2v$
• B. $1.08v$
• C. $1.02v$
• D. $1.5v$

Answer 1

The correct option is C $1.02v$

Given:
Initial root mean square speed, $v_1=v$
Initial temperature of the gas, $T_1={20}^{\circ }\text{C}=293\text{K}$
Change in temperature, $\Delta T=10{}^{\circ }\text{C}$
So, final temperature of the gas, $T_2={30}^{\circ }\text{C}=303\text{K}$
To find:
Final root mean square speed, $v_2=?$

We know that, root mean square speed is given by,
$v_{rms}=\sqrt{\frac{3RT}{M}}$
Here, $R$ is the gas constant, $T$ is the temperature and $M$ is the molar mass.
So, $v_1=\sqrt{\frac{3R\times 293}{M}}$ and
$v_2=\sqrt{\frac{3R\times 303}{M}}$
[ gas is same, so molar mass is same ]
Now, $\frac{v_2}{v_1}=\sqrt{\frac{303}{293}}$
$\Rightarrow v_2\approx 1.02v_1\approx 1.02v$