Question

A 55.0 L steel tank at 20.0 ∘C contains acetylene gas, C2H2, at a pressure of 1.39 atm. Assuming ideal behavior, how many grams of acetylene are in the tank?

Answer 1

Answer:

PV=nRT

n = PV/RT

n = m/Mm

m/Mm = PV/RT

m = MmPV/RT

T in kelvin = T Celsius + 273.15 = 293.15 K

m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)

mass in grams = 82.8 grams  

Explanation:

Ideal gases formula is PV=nRT, where:

P is the pressure (1.39 atm in this case)

V is the volume (55.0 L in this case)

R is the gas constant (0.08206 L.atm/K.mole)

T is the temperature (20.0C) should be converted to Kelvin

all the unit should correspond to the one in the R.

we also know that to find the mass, we can use number mole with the formula number of mole(n) = mass (m) divided by the molar mass (Mm). therefore we substituted that in the formula and make (m) the subject of the formula.

we found the mass to be 82.8 grams

Answer 2

PV=nRT is the formula for ideal gases, where P is the pressure and R is the constant and T is the temperature (1.39 atm in this case)

• The letter V stands for volume (55.0 L in this case)
• (0.08206 L.atm/K.mole) is the gas constant.
• T stands for temperature (20.0C) All units should be changed to Kelvin and correspond to the one in the R.
• We also know that we can apply the formula number of mole(n) = mass (m) divided by the molar mass to obtain the mass (Mm). As a result, we inserted that in the formula and made (m) the subject.

$m = MmPV/RT\\m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)$

Thus, the mass was discovered to be 82.8 grams.

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