Question

A cup of boiling noodle water was vaporized into gas. The entire amount of water vapor was 200 grams, 400 K, and 1.00 L. What was the pressure in atm?

Answer 1

The pressure of the water vapor gas is 364.300 atm

To solve this question, we need to understand the concept of Ideal gas law.

What is Ideal Gas Law?

The ideal gas law has a relationship between the pressure, temperature, and volume of gases, and it is also a good approximation for the characteristics of gases at different conditions.

It is given by the formula:

• PV = nRT

where;

• n = number of moles of the gas

From the parameter given:

• The mass of water vapor = 200 grams

We know that the molar mass of water vapor at standard conditions is 18.02 g/mol

Thus, we can calculate the number of moles of water vapor by using the expression:

number of moles (n) = mass/molar mass

$\mathbf{ n= \dfrac{200 g}{18.02 \ g/mol} }$

n = 11.099 moles

From the ideal gas equation:

PV = nRT

$\mathbf{P = \dfrac{nRT}{V}}$

$\mathbf{P = \dfrac{11.099 \ moles \times0.082057 L.atm/K/mol \times 400 \ K }{1.00 \ L}}$

P = 364.300 atm

Learn more about the Ideal gas equation here:

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