Respuesta :
By using the ideal gas formula:
1) TO get the volume: from the ideal gas formula:
when PV = nRT
when we have P = 1 & R = 0.0821 & T= 0°C+273 = 273 K
and n = mass / molar mass of He
= 8 g / 4 g/mol = 2 mol
so by substitution:
1 atm * V = 2 mol * 0.0821 * 273
∴V = 44.8 L
2) To get moles of gas molecules we will use also the ideal gas formula:
PV = nRT
when P = 1 atm & V=2 L & R = 0.0821 & T = 0+273 = 273 K
so by substitution:
1atm * 2 L = n * 0.0821 * 273
∴n (the no.of moles)= 0.089 mol
to get the no.of moles of gas molecules we will multiply by avogadro's number:
= 0.089 * (6.02*10^23) = 5.37X10^22 molecules
3) first we have to get n:
PV = n RT
when P= 1 atm & V = 3L & R= 0.0821 & T = 0+273 = 273 K
so by substitution:
1atm * 3L = n * 0.0821 * 273
∴n the no.of moles of the gas= 0.134 mol
a) when the gas is He so the mass can be calculated from this formula:
He Mass = no.of moles * molar mass of He
= 0.134m * 4g/mol
= 0.5 g
b) when the gas is CO2 So the mass can be calculated from this formula also:
CO2 Mass = no.of moles * molar mass of CO2
=0.134mol * 44 g/mol
= 5.896 g
c) when the gas is CH4 So the mass can be calculated from this formula :
CH4 Mass = no.of moles * molar mass of CH4
= 0.134mol * 16 g /mol
= 2.144 g
4) by using also the ideal gas formula we can get the moles of O2:
PV = nRT
when P= 1 atm & V = 0.001 L & R = 0.0821 & T = 0+273 = 273 K
so by substitution:
1atm * 0.001 L = n * 0.0821 *273
∴ n = 4.5X10^-5 mol
1) TO get the volume: from the ideal gas formula:
when PV = nRT
when we have P = 1 & R = 0.0821 & T= 0°C+273 = 273 K
and n = mass / molar mass of He
= 8 g / 4 g/mol = 2 mol
so by substitution:
1 atm * V = 2 mol * 0.0821 * 273
∴V = 44.8 L
2) To get moles of gas molecules we will use also the ideal gas formula:
PV = nRT
when P = 1 atm & V=2 L & R = 0.0821 & T = 0+273 = 273 K
so by substitution:
1atm * 2 L = n * 0.0821 * 273
∴n (the no.of moles)= 0.089 mol
to get the no.of moles of gas molecules we will multiply by avogadro's number:
= 0.089 * (6.02*10^23) = 5.37X10^22 molecules
3) first we have to get n:
PV = n RT
when P= 1 atm & V = 3L & R= 0.0821 & T = 0+273 = 273 K
so by substitution:
1atm * 3L = n * 0.0821 * 273
∴n the no.of moles of the gas= 0.134 mol
a) when the gas is He so the mass can be calculated from this formula:
He Mass = no.of moles * molar mass of He
= 0.134m * 4g/mol
= 0.5 g
b) when the gas is CO2 So the mass can be calculated from this formula also:
CO2 Mass = no.of moles * molar mass of CO2
=0.134mol * 44 g/mol
= 5.896 g
c) when the gas is CH4 So the mass can be calculated from this formula :
CH4 Mass = no.of moles * molar mass of CH4
= 0.134mol * 16 g /mol
= 2.144 g
4) by using also the ideal gas formula we can get the moles of O2:
PV = nRT
when P= 1 atm & V = 0.001 L & R = 0.0821 & T = 0+273 = 273 K
so by substitution:
1atm * 0.001 L = n * 0.0821 *273
∴ n = 4.5X10^-5 mol