A gas sample in a rigid container at 455 K is brought to STP (273K and 1 atm). What was the original pressure of the gas in mmHg

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Chemistry

Quinn Glenn

26 June, 21:58

A gas sample in a rigid container at 455 K is brought to STP (273K and 1 atm). What was the original pressure of the gas in mmHg

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Alexa Riley · Answer 1

1) V (CH₄) = 0,376 L. T (CH₄) = 304 K. p (CH₄) = 1,5 atm 101325 Pa/atm = 151987,5 Pa = 151,9875 kPa. R = 8,314 J/K·mol. Use ideal gas law: p·V = n·R·T. n (CH₄) = p · V : R · T. n (CH₄) = 151,9875 kPa · 0,376 L : 8,314 J/K· mol · 304 K. n (CH₄) = 0,0226 mol. V (CH₄) = n (CH₄) · Vm. V (CH₄) = 0,0226 mol · 22,4 dm³/mol. V (CH₄) = 0,506 dm³ = 0,506 L.

2) V (SO₂) = 5,2 L. p (SO₂) = 45,2 atm = 45,2 atm · 101,325 kPa/atm = 4579,89 kPa. T (SO₂) = 293 K. R = 8,314 J/K·mol. Use ideal gas law: p·V = n·R·T. n (SO₂) = p · V : R · T. n (SO₂) = 4579,89 kPa · 5,2 L : 8,314 J/K· mol · 293 K. n (CH₄) = 9,77 mol. There is not enogh SO₂, 225 mol - 9,77 mol = 215,23 mol is needed.

3) p (He) = 3,50 atm · 101,325 kPa/atm = 354,63 kPa. V (He) = 4,00 L. n (He) = 0,410 mol. R = 8,314 J/K·mol. Use ideal gas law: p·V = n·R·T. T = p · V : R · n. T (He) = 354,63 kPa · 4,00 L : 8,314 J/K· mol · 0,410 mol. T (He) = 416,14 K. n - amount of substance.

4) p (Ar) = 1,00 atm · 101,325 kPa/atm = 101,325 kPa. V (Ar) = 3,4 L. T (Ar) = 263 K. R = 8,314 J/K·mol. Use ideal gas law: p·V = n·R·T. n (Ar) = p · V : R · T. n (Ar) = 101,325 kPa · 3,4 L : 8,314 J/K· mol · 263 K. n (Ar) = 0,157 mol. n (Ar) = 0,157 mol + 2,5 mol = 2,657 mol. p (Ar) = 2,657 mol · 8,314 J/K· mol · 263 K : 3,4 L. p (Ar) = 1708,74 kPa.