Question

The pressure exerted by 1.30 mol of gas in a 13.0 L container at 295 K is

Answer 1

The pressure of the gas is 2.42atm.

Step-by-step explanation:

The given information from the exercise is:

- Number of moles (n): 1.30 moles

- Volume (V): 13.0L

- Temperature (T): 295K

With the Ideal Gases Law formula we can calculate the pressure (P) of the gas, by replacing the values of n, V and T:

`\begin{gathered} P*V=n*R*T \\ P*13.0L=1.30mol*0.082(atm*L)/(mol*K)*295K \\ P*13.0L=31.45atm*L \\ P=(31.45atm*L)/(13.0L) \\ P=2.42atm \end{gathered}`

So, the pressure of the gas is 2.42atm.