The answer is rounded to two sig figs, the number of sig figs you have for the volume of the gas. What is the total pressure (in atm) when the valve is opened 18. The right hand cylinder contains 3.0 L of He at 0.60 atm and the left hand cylinder contains 6.8 L of He at 2.0 atm. Assume two cylinders at 27 вЊ are connected by a closed valve system. #m = overbrace((PV)/(RT))^(color(blue)('the number of moles')) * M_M#ДЇinally, plug in your values to find the mass of the sample-do not forget to convert the temperature from degrees Celsius to Kelvin and the pressure from mmHg to atmospheres!

This means that you can rewrite the ideal gas law equation as What will the volume of the gas be at a pressure of 1. You also know that the mass of the gas, let's say #m#, can be expressed in terms of its molar mass, let's say #M_M#, and the number of moles. A sample of gas has a volume of 250 cm3 when its pressure is 746 mmHg. #T# is the absolute temperature of the gas.This is done in order to convert the pressure from mmHg to atm, because the value for R contains atm as the pressure unit. #n# is the number of moles of gas present in the sample.Standard atmospheric pressure of 1013.25 mb at 0 m, 226.321 mb at 11,000 m, 54.7489 mb at 20,000 m above mean sea level.

#color(blue)(ul(color(black)(PV = nRT)))# The height relative to sea level conversions are derived using the US Standard Atmosphere 1976 barometric formula for geopotential altitude with the following values1-6 at heights from 5,000 ft (1524 m) below to 100,000 ft (30,480 m) above mean sea level. So, you know that the ideal gas law equation looks like this

You could also rewrite the ideal gas law equation in terms of the mass of helium present in the sample to get the answer directly. The idea here is that you can use the ideal gas law equation to find the number of moles of helium present in your sample and the molar mass of the gas to convert that to grams.

746 mmhg to atm