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?

Question

Jerimiah Ford

Chemistry

25 July, 16:08

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?

+3

Answers (1)

Know the Answer?

Heath Carey · Answer 1

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