Question

What pressure, in kPa, will 0.416 mol of krypton gas occupy at standard temperature and 45.6 L?

Answer 1

Answer: the pressure of the gas under the conditions given is 20.7kPa

Step-by-step explanation:

The question requires us to determine the pressure of a gas, in kPa, knowing that there are 0.416 moles of this gas occupy 45.6 L of volume at standard temperature.

We can apply the rearranged equation for ideal gases to solve this problem, as shown below:

`P* V=n* R* T\rightarrow P=(n* R* T)/(V)`

where P is the pressure of the gas, V is its volume (45.6 L), n is the number of moles of gas (0.416 mol), T is the temperature (standard temperature) and R is the constant of gases (we'll apply the value 8.314 kPa.L/mol.K).

The standard temperature can be defined as 273.15 K (or 0°C), as stated by the Standard Temperature and Pressure conditions (STP).

Applying the values provided by the question, we'll have:

`\begin{gathered} \begin{equation*} P=(n* R* T)/(V) \end{equation*} \\ \\ P=(0.416\text{ mol\rparen}*(8.314\text{ kPa.L/mol.K\rparen}*(273.15\text{ K\rparen}*\frac{1}{(45.6\text{ L\rparen}} \\ \\ P=3.46\text{ \lparen kPa.L/K\rparen}*(273.15\text{ K\rparen}*\frac{1}{(45.6\text{ }L)} \\ \\ P=(944.72\text{ kPa.L\rparen}*\frac{1}{(45.6\text{L})} \\ \\ P=20.7\text{ kPa} \end{gathered}`

Therefore, the pressure of the gas under the conditions given is 20.7kPa.