What volume of ar at 53.6 °c and 1.31 atm contains the same number of particles as 0.584 l of h2 at 19.6 °c and 2.10 atm?

Volume = 1.043L

Explanation:

First off, we have to calculate the number of particles H2 at that condition have.

We were given the following;

Volume (V) = 0.584L

Temperature = 19.6°C + 273 = 292.6 K (Converting to Kelvin Temperature)

Pressure = 2.10 atm

Using the ideal gas equation;

PV = nRT,

where R = gas constant = 0.0821 L atm K−1 mol−1

making n subject of formular;

n = PV / RT

n = (2.1 * 0.584) / (0.0821 * 292.6)

n = 1.2264 / 24.02246

n = 0.05098 moles

Avogadro’s number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance. It is equal to 6.022×1023 mol-1

Number of particles present in 0.05098 moles of H2 is given as;

1 mole = 6.022×1023

0.05098 = x

Upon cross multiplication, we have;

x = 0.05098 * 6.022×1023 = 0.307×1023 = 3.07×1022 particles.

equal number of moles would have the same number of particles. So we just have to find the volume when the number of moles of Ar is 0.05098.

We were given the following for Ar;

Volume (V) = ?

Temperature = 53.6°C + 273 = 326.6 K (Converting to Kelvin Temperature)

Pressure = 1.31 atm

Using the ideal gas equation;

PV = nRT,

where R = gas constant = 0.0821 L atm K−1 mol−1

making V subject of formular;

V = nRT / P

V = ( 0.05098 * 0.0821 * 326.6) /  1.31

V = 1.043L

yes the answer is total mass in unchanged   : )