Question

if you have 32 grams of water vapor in a bottle that holds 1.5 L what is the pressure at standard temperature?

Answer 1

26.58 atm

Step-by-step explanation:

By the ideal gas law, we have that:

`PV=\text{nRT}`

Where P is the pressure, V is the volume, n is the number of moles, R is a constant equal to 0.08206 L atm/ mol K, and T is the temperature.

If we have 32 grams of water vapor, the number of moles will be equal to:

`32\text{ grams }*\frac{1\text{ mol}}{18.01\text{ gr}}=1.78\text{ moles}`

Because 18.01 gr is the molar mass of the vapor water.

On the other hand, the standard temperature is 273 K. So, replacing the values for each constant, we get:

`\begin{gathered} PV=\text{nRT} \\ P(1.5L)=(1.78\text{ mol)}(0.08206\text{ }\frac{L\text{ atm}}{mol\text{ K}})(273\text{ K)} \\ P(1.5\text{ L) = 39.87 L atm} \end{gathered}`

So, dividing both sides by 1.5 L:

`\begin{gathered} (P(1.5L))/(1.5L)=\frac{39.87\text{ L atm}}{1.5} \\ P=26.58\text{ atm} \end{gathered}`

Therefore, the pressure is 26.58 atm.