Calculate the density of N_2 at STP. The density of an unknown ideal gas at 20.0
ID: 511622 • Letter: C
Question
Calculate the density of N_2 at STP. The density of an unknown ideal gas at 20.0 degree C and 1.00 atm was found to be 1.83 g/L. What is the molar mass of the unknown? Calculate the pressure exerted by 65.0 kg of Ar in a 756.0-L tank at 30.0 degree C. A tank contains 105 g of O_2 at 35 degree C and 860 mmHg. Calculate the new volume after the removal of 30.0 g of oxygen under the same conditions. A 25.0-L balloon at 3.0 atm and 25 degree C contains 12.0 g of He. How many grams of He will inflate the balloon twice its initial size under the same conditions?Explanation / Answer
(6) At STP,
Pressure, P = 1 atm
Temperature, T = 273.15 K
Gas constant, R = 0.0821 L.atm.K-1.mol-1
Molar mass of N2, M = 28 g/mol
Assuming ideal gas behaviour for nitrogen gas,
Ideal gas equation,
P V = n R T
But, number of moles, n = mass / molar mass = m / M
So, P V = (m/M) R T
P M = (m/V) R T
But, density, d = mass / volume = m/V
SO, P M = d R T
d = P M / R T
d = 1 * 28 / (0.0821 * 273.15)
d = 1.25 g/L
(7)
Given that,
Density, d = 1.83 g/L
Temperatue, T = 20.0 + 273.15 = 293.15 K
Pressure, P = 1.00 atm
R = 0.0821 L.atm.K-1.mol-1
M = ?
We know tha,
d = P M / R T
(OR)
M = d R T / P
M = 1.83 * 0.0821 * 293.15 / 1.00
M = 44.0 g/mol
(8)
Given that,
Mass of Ar = 65.0 kg. = 6.50 * 104 g.
Molar mass of Ar = 40.0 g/mol
Number of moles of Ar, n = mass / molar mass = 6.50 * 104 / 40.0 = 1.62 * 103 mol
Volume, V = 756.0 L
Temperature, T = 30.0 + 273.15 = 303.15 K
R = 0.0821 l.atm.K-1.mol-1
Ideal gas equation,
P V = n R T
P = 1.62 * 103 * 0.0821 * 303.15 / 756.0
P = 53.3 atm
(9)
Initial Mass of O2 = 105 g.
molar mass of O2 = 32 g/mol
Initial number of moles of O2, n1 = 105/32 = 3.28 mol
Temperautre, T = 35 + 273.15 = 308.15 K
Pressure, P = 860 mmHg = 860 / 760 = 1.13 atm
R = 0.0821 L.atm.K-1.mol-1
Assuming ideal behaviour,
P V1 = n1 R T
V1 = 3.28 * 0.0821 * 308.15 / 1.13
V1 = 73.4 L
final mass of O2 = 105 - 30.0 = 75.0 g.
final number of moles of O2, n2 = 75 / 32 = 2.34 mol
V2 = final volume = ?
According to Avagadro's law of ideal gases,
V1 / n1 = V2 / n2
73.4 / 3.28 = V2 / 2.34
V2 = final volume = 52.4 L
(10)
Initial volume, V1 = 25.0 L
Initial moles of He, n1 = 12.0 /4 = 3.00 mol
Final volume, V2 = 2 * 25.0 = 50.0 L
Final number of moles, n2 = ?
According to Avagadro's law,
V1 / n1 = V2 / n2
25.0 / 3.00 = 50.0 / n2
n2 = final number of moles = 6.00 mol
There, final mass of He = 6.00 * 4 = 24.0 g.