part A part B and part C

Question

part A

0.755 mol sample of argon gas at a temperature of 20.0 °C is found to occupy a volume of 22.9 liters. The pressure of this gas sample is  mm Hg.

part B

A sample of methane gas collected at a pressure of 520 mm Hg and a temperature of 292 K has a mass of 23.7 grams. The volume of the sample is  L.

part C

A 0.701 gram sample of hydrogen gas has a volume of 979 milliliters at a pressure of 2.84 atm. The temperature of the H₂ gas sample is  °C.

Answer 1

All of these questions are asking you to apply the Ideal Gas Law, PV = nRT, however parts B and C you first need to convert the mass into moles.

Part A

P (22.9 L) = 0.755 mol (62.36 L mmHg/mol K)(293 K) ---- here I used 62.36 because the question is asking for the pressure in mmHg.

P = 602 mmHg

Part B

23.7 g CH₄ (1 mol CH₄ / 16.042 g CH₄) = 1.477 mol CH₄

(520 mmHg)(V) = 1.477 mol (62.36 L mmHg/mol K)(292 K)

V = 52 L

Part C

0.701 g H₂ (1 mol H₂/ 2.016 g H₂) = 0.3477 mol H₂

979 mL = 0.979 L

2.84 atm (0.979 L) = 0.3477 mol (0.08206 L atm/mol K)(T) ----- here I used 0.08206 because now the pressure is in atm.

T = 97.4 K

T = 97.4 - 273 = -176 ^oC