A 0.0434-m3 container is initially evacuated. Then, 4.19 g of water is placed in the container, and, after some time, all of the water evaporates. If the temperature of the water vapor is 417 K, what is its pressure? Number units in Pa

P = 18760.5 Pa

Explanation:

Given that

Volume, V = 0.0434 m³

Mass, m = 4.19 g = 0.00419 kg

T = 417 K

If we assume that water vapor is behaving like a ideal gas, then we can use ideal gas equation

Ideal gas equation P V = m R T

p=Pressure, V = Volume, m = mass

T=Temperature, R=Universal gas constant

Now by putting the values

P V = m R T

For water R = 0.466 KJ/kgK

P x 0.0434 = 0.00419 x 0.466 x 417

P = 18.7605 KPa

P = 18760.5 Pa

Therefore the answer is 18760.5 Pa