Question

What pressure (in kilopascals) is exerted by 4.20 moles of Xenon gas in a 15.0 L container at 280.0 K?

Answer 1

Answer: the pressure exerted by the gas is 652 x 10^3 Pa, which corresponds to 652 kPa

Explanation:

The question requires us to calculate the pressure, in kPa, connsidering the following information:

number of moles = n = 4.20mol

volume of gas = V = 15.0L

temperature of gas = T = 280.0 K

We can use the equation of ideal gases to calculate the pressure of the gas, as shown by the rearranged equation below:

$PV=nRT\rightarrow P=\frac{nRT}{V}$

Since the volume was given in L and the question requires us to calculate the pressure in kPa, we can use R in units of L.Pa/K.mol:

R = 8314.46 L.Pa/K.mol

Applying the values given by the question to the rearranged equation above, we'll have:

$\begin{gathered} P=\frac{nRT}{V} \\ \\ P=\frac{(4.20mol)\times(8314.46L.Pa/K.mol)\times(280.0K)}{(15.0L)}=652\times10^3Pa \end{gathered}$

Therefore, the pressure exerted by the gas is 652 x 10^3 Pa, which corresponds to 652 kPa.