Question #624ba

1 Answer
Mar 17, 2016

"3.00 moles"

Explanation:

The idea here is that one mole of any ideal gas that is kept under Standard Temperature and Pressure, STP, conditions will occupy "22.4 L " -> this is known as the molar volume of a gas at STP.

This can be derived using the ideal gas law equation

color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" ", where

P - the pressure of the gas
V - the volume occupied by the gas
n - the number of moles of gas present
R - the universal gas constant, usually given as 0.0821("atm" * "L")/("mol" * "K")
T - the absolute temperature of the gas

STP conditions are defined as a pressure of "1 atm" and a temperature of "273.15 K". Plug these values into the ideal gas law equation and solve for V/n

PV = nRT implies V/n = (RT)/P

V/n = (0.0821(color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * 273.15color(red)(cancel(color(black)("K"))))/(1color(red)(cancel(color(black)("atm")))) = "22.4 L mol"^(-1)

This tells you that every mole of an idea lgas that is being kept under STP conditions will occupy "22.4 L".

http://slideplayer.com/slide/1716615/http://slideplayer.com/slide/1716615/

So, if "22.4 L" correspond to one mole of any ideal gas, it follows that "67.2 L" will correspond to

67.2color(red)(cancel(color(black)("L"))) * overbrace("1 mole"/(22.4color(red)(cancel(color(black)("L")))))^(color(purple)("molar volume of a gas at STP")) = color(green)(|bar(ul(color(white)(a/a)"3.00 moles"color(white)(a/a)|)))

SIDE NOTE It's worth noting that the current STP conditions are defined as a pressure of "100 kPa" and a temperature of "273.15 K".

Under these conditions for pressure and temperature, one mole of any ideal gas occupies "22.7 L".