Question

According to Maxwell's distribution of molecular speeds, for the below graph drawn for two different samples of gases A and B at temperature $T_1$ and $T_2$ respectively, which of the following statements is/are correct?

• A. If $T_1=T_2$, then molecular mass of gas $B\left(M_B\right)$ is less than molecular mass of gas $A\left(M_A\right)$
• B. If molecular mass of gas $A\left(M_A\right)$ is equal to molecular mass of gas $B\left(M_B\right)$, then $T_1<T_2$
• C. If $T_1<T_2$, then molecular mass of gas $B\left(M_B\right)$ is necessarily more than molecular mass of gas $A\left(M_A\right)$
• D. If gas A is $O_2$ and gas B is $N_2$, then $T_1$ is necessarily more than $T_2$

Answer 1

Answer: (A), (B)

The correct options are
A If $T_1=T_2$, then molecular mass of gas $B\left(M_B\right)$ is less than molecular mass of gas $A\left(M_A\right)$
B If molecular mass of gas $A\left(M_A\right)$ is equal to molecular mass of gas $B\left(M_B\right)$, then $T_1<T_2$

$V_{mp}=\sqrt{\frac{3RT}{M}}$
Most probable speed is represented by x-cordinate of peak of maxwell distribution. Since, x-cordinate of peak of curve B is more, therefore, most probable speed of B is more

At constant temperature, most probable speed is inversely propotional to molecular mass. Hence, if $T_1=T_2$, then molecular mass of gas $B\left(M_B\right)$ is less than molecular mass of gas $A\left(M_A\right)$.

At constant molecular mass, most probable speed is directly propotional to temperature. Since,most probable speed of B is more, therefore, temperature of B is more. Hence, if molecular mass of gas $A\left(M_A\right)$ is equal to molecular mass of gas $B\left(M_B\right)$, then $T_1<T_2$.

Since, most probable speed of B is more,therefore, $\frac{T_1}{M_A}<\frac{T_2}{M_B}$ & $\frac{T_2}{T_1}>\frac{M_B}{M_A}$. Hence, if $T_1<T_2$, then molecular mass of gas $B\left(M_B\right)$ can be more or equal or less than molecular mass of gas $A\left(M_A\right)$ as in all case the inequality can hold.

Applying same logic, as $\frac{T_2}{T_1}>\frac{M_B}{M_A}$ if molecular mass of gas $A\left(M_A\right)$ is more than molecular mass of gas $B\left(M_B\right)$ as gas A is oxygen whose molecular mass is more than gas B which is nitrogen , then $T_1$ can be less or equal or more than $T_2$ as in all case inequality will hold.