Question

There are 215.0 mol of an unknown gas stored at a temperature of 1110 K in a large, 395 L container. R=0.0821 L·atm/(mol·K) In atm, what is the approximate pressure of the gas?

Answer 1

To solve this problem we use the ideal gas equation:

`P.V=n.R.T`

Where:

P is the pressure of the gas

V is the volume of the container (395 L)

n is the number of moles present (215 mol)

R is the ideal gas constant (0.0821 L.atm/mol.K)

T is the temperature of the gas (1110 K)

So to calculate the pressuere we have to use the equation and the data:

`\begin{gathered} P=(n.R.T)/(V) \\ P=\frac{215mol\text{ . }0.0821(L.atm)/(mol.K)\text{ . }1110K}{395L} \\ P=\text{ 49.6 atm} \end{gathered}`

So the answer is 49.6 atm.