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College Physics (Urone)
Found in: Page 472

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Short Answer

Question: The vapor pressure of water at 40.0ºC is 7.34×103 N/m2. Using the ideal gas law, calculate the density of water vapor in g/m3 that creates a partial pressure equal to this vapor pressure. The result should be the same as the saturation vapor density at that temperature (51.1 g/m3).


The density of water vapor is 50.746 g/m3.

See the step by step solution

Step by Step Solution

Step 1: Defining the ideal gas law

Ideal gas law states that the product of pressure and volume of a gas is equal to the product of temperature in Kelvin and the universal gas constant. It is expressed as;


Where is the pressure, is the volume, is the number of moles,is the universal gas constant, and is the temperature in Kelvin.

Step 3: Evaluate the given factors and convert them into suitable units

The value of pressure and temperature are given.

P=7.34×103 N/m3T=40.0C=40+273.15K=313.15 K

The value of is 8.314 Jmol-1K-1. No of moles is the ratio of a given mass of a substance to the molar mass. It is expressed as;

n=MassMolar Mass……..…(2)

The molar mass of water is 18g. Substitute this in equation (2).

n=Mass18 g…………….(3)

Step 3: Find density from the ideal gas formula

The density of water vapor can be derived from the ideal gas formula.

Substitute equation (3) in equation (1)

PV=Mass18 gRTMassV=P×18 gRT

The ratio of mass to volume is the density. Therefore,

Density=P×18 gRT =7.34×103 N/m2×18g8.314 Jmol-1K-1×313.15 K =50.746g/m3

This value is approximately equal to 51.1 g/m3, which is the saturation vapor density at the given temperature. The small variation is due to the error caused by rounding the values.

So, the calculated vapor density is equal to 50.746 g/m3.

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