Question

2.0 g of helium gas were placed in a glass container at 25°C. The pressure was measured to be 120 mmHg. If 6.0 g of He gas were added to the container while the temperature constant, how much would the pressure be?

Answer 1

: 1.09 g of H2 is contained in a 2.00 L container at 20.0 °C. What is the pressure in this container in mmHg?

Solution:

Rearrange the Ideal Gas Law to this:

P = nRT / V

Substitute values into the equation:

P = [(1.09 g / 2.02 g mol¯1) (0.08206 L atm mol¯1 K¯1) (293.0 K)] / 2.00 L

Please note the division of 1.09 by 2.02. This is done in order to convert grams to moles, because the value for R contains mol as the unit for amount of substance. If we used g, the mol unit in R would not cancel and we need to have it cancel because we require atm (and only atm) to be in the answer.

Multiply the answer (which is in atm) by 760.0 mmHg atm¯1 to get mmHg