Find the temperature at which the rate of effusion of N₂ is 1.625 times to that of SO₂ at 500℃

The correct option is 620℃.

According to Graham's law of effusion, the rate of effusion of a gas ($r$) is directly proportional to its temperature in Kelvin ($T$) and inversely proportional to the square root of its molar mass ($M$).
This relationship can be expressed mathematically as:
$r\propto \sqrt{\frac{T}{M}}$

Let us write down the given parameters and constants:
For nitrogen gas (N₂):
Molar mass, $M_{{\text{N}}_2}=28\text{g/mol}$
Let its temperature be $T_{{\text{N}}_2}$ in Kelvin.
For sulfur dioxide gas (SO₂):
Molar mass, $M_{{\text{SO}}_2}=32+2\left(16\right)=64\text{g/mol}$
Temperature, $T_{{\text{SO}}_2}=500℃=500+273=773\text{K}$

The ratio of the rate of effusion of N₂ to that of SO₂ is:
$\frac{r_{{\text{N}}_2}}{r_{{\text{SO}}_2}}=\sqrt{\frac{T_{{\text{N}}_2}}{T_{{\text{SO}}_2}}\times \frac{M_{{\text{SO}}_2}}{M_{{\text{N}}_2}}}$

Given that the rate of effusion of N₂ is 1.625 times that of SO₂:
$\frac{r_{{\text{N}}_2}}{r_{{\text{SO}}_2}}=1.625$

Substitute the values into the equation:
$1.625=\sqrt{\frac{T_{{\text{N}}_2}}{773}\times \frac{64}{28}}$

Squaring both sides to remove the square root:
${\left(1.625\right)}^2=\frac{T_{{\text{N}}_2}}{773}\times \frac{16}{7}$
$2.640625=T_{{\text{N}}_2}\times \frac{16}{5411}$

Solving for $T_{{\text{N}}_2}$:
$T_{{\text{N}}_2}=\frac{2.640625\times 5411}{16}$
$T_{{\text{N}}_2}\approx 893\text{K}$

Convert the temperature from Kelvin back to Celsius:
$\text{Temperature in}℃=893-273=620℃$

Therefore, the temperature at which the rate of effusion of N₂ is 1.625 times that of SO₂ is 620℃.