Question

What pressure (in kilopascals) is exerted by 3.70 moles of Xenon gas in a 12.0 L container at 290.0 K?

Answer 1

The pressure of the gas is 743.4 Kpa

Step-by-step explanation

Given that;

The volume of the gas is 12L

The number of moles of the gas 3.70 moles

The temperature of the gas is 290K

Follow the steps below to find to the pressure of the gas

The ideal gas equation is given below as

`\text{ PV = nRT}`
`\begin{gathered} \text{ R = 8.314 L.Kpa. mol}^(-1)K^(-1) \\ \text{ P }*\text{ 12 = 3.70}*\text{ 290}*\text{ 8.314} \\ \text{ 12P = 8290.922} \\ \text{ Divide both sides by 12} \\ \text{ P = }\frac{\text{ 8290.922}}{\text{ 12}} \\ \text{ P = 743.4 Kpa} \end{gathered}`

Therefore, the pressure of the gas is 743.4 Kpa