Question

At what temperature in Celsius will $19.4{\text{ }}g$ of molecular ozone exert a pressure of $1820{\text{ }}mmHg$ in a $5.12{\text{ }}L$ cylinder?

Answer 1

This is a question of an ideal gas law. In the question pressure, volume and mass of the molecular ozone is given so using that you can calculate the temperature. Here the number of moles is not given but mass is given so it can be converted to moles if molar mass of the molecule is known. Use the formula: $PV = nRT$
(Where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in Kelvins).

Complete Step By Step Answer:
In the question we are given the weight of molecular oxygen $= 19.4g$ , pressure exerted $= 1820{\text{ }}mmHg$ and volume of the cylinder $= 5.12L$ , and we need to find the temperature. For solving this question we have to use the ideal gas law.
The formula for ideal gas law is:
$PV = nRT$
Where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in Kelvins.
In the question number of moles of ozone is not given but we are given with the weight of the ozone so we convert this weight in grams into moles by using the formula:
$Mass \times \dfrac{{1{\text{ }}mol{\text{ }}of{\text{ }}{O_3}}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}{O_3}}}$ where molar mass of ozone is $48\;gmo{l^{ - 1}}$
$\Rightarrow 19.4g \times \dfrac{{1{\text{ }}mol{\text{ }}of{\text{ }}{O_3}}}{{48g{\text{ }}{O_3}}}$
$\Rightarrow 0.404{\text{ }}mol$
Thus the number of moles of ozone is $0.404{\text{ }}mol$
Now that we have all the values we will find the temperature. the formula to be used is:
$PV = nRT$
The value of gas constant is $62.3635{\text{ }}LmmHg{K^{ - 1}}mo{l^{ - 1}}$
We will rearrange the formula for finding the value of T. the new formula is:
$T = \dfrac{{PV}}{{nR}}$
By substituting the values we get:
$T = \dfrac{{1820 \times 5.12}}{{0.404 \times 62.3635}}$
$\Rightarrow T = \dfrac{{9318.4}}{{25.194854}}$
$\Rightarrow T = 369.85 \approx 370K$
Therefore the temperature is $370{\text{ }}K$ .

Note:
While solving the problem we have taken the value of the gas constant as $62.3635{\text{ }}LmmHg{K^{ - 1}}mo{l^{ - 1}}$ . We have taken this value because in the question the pressure is given in terms of $mmHg$ and volume is given in liters. But if the pressure was given in terms of atmosphere and volume in milliliters, then the value of gas constant will be taken as $82.1{\text{ }}mL{\text{ }}atm{\text{ }}{K^{ - 1}}mo{l^{ - 1}}$ .