Many laboratory gases are sold in steel cylinders with a volume of 43.8 l. what mass (in grams) of argon is inside a cylinder whose pressure is 17615kpa at 23∘c?

It's very simple... if we remember value of Universal Gas Constant R and Ideal Gas Law, so...

Ideal Gas Law
pV = nRT, where:
p - pressure (in kPa),
V - volume (in L),
n - number of moles (in mol),
R - universal cas constant (in kPa * L / mo l* K),
T - temperature (in K)

n = m/M, where:
n - number of moles,
m - mass (in grams),
M - molar mass of ingredient (in g/mol) - you find this at Periodic Table.

pV = nRT ---> pV = mRT/M ---> pVM = mRT ---> pVM/RT = m

p = 17615 kPa
T = 273.15 + 23 = 296.15 K
V = 43.8 L
R = 8.314 kPa * L / mol * K
M (for argon) = 39.948 g/mol

and

m = (17615 kPa * 48.3 L * 39.948 g/mol) / (296.15 K * 8.314 kPa * L / mol * K)
m = 13803.93 grams of Argon

Particle exchange energy though elastic collision , hope this .