A sample of oxygen gas at 791 torr and 25degree C occupies a volume of 20.0 L Ho

Question

A sample of oxygen gas at 791 torr and 25degree C occupies a volume of 20.0 L How many liters would it occupy at 461 torr and 25 degree C? A sample of 4.92 Times 10^10 atoms of nitrogen gas is at a temperature of 30.0 degree C. How many molecules of nitrogen gas would have to be added to the sample, at constant pressure and volume, to lower the temperature of the gas to 25.0 degree c? Calculate the density of helium gas at 1.07Times10^-6 torr and -252 degree C. A gaseous compound is analyzed and found to contain 6.67% hydrogen, 40.0% carbon, and 53.33% oxygen by mass. The gaseous compound has a density of 0.964 g/L at 25.2 degree C and 598 torr. Find the molecular formula of the gaseous compound. A gas mixture contains 15.0 g of carbon dioxide, 15.0 g of nitrogen, and 15.0 g of helium. Calculate the mole fraction of helium in the mixture.

Explanation / Answer

1)

Given data:

Initial conditions be, P1 = 791 torr, V1 = 20.0 L

Finally, P2 = 461 torr, V2 =?

Temperature is constant here,

This question is based on Boyle’s law, state mathematically as,

P 1/V       ……. (at T = constant)

Hence, PV = constant.

i.e. P1V1 = P2V2 = …. = constant,

P2V2 = P1V1

Let us put known values and solve the equation,

461 x V2 =791 x 20

V2 = 15820/461

V2 = 34.32 L

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2)

Given data:

Initial : n1 = 4.92 x 1010 T1 = 30 oC = 30 + 273 = 303 K

Final: n2 = ? T2 = 25 oC = 25 + 273 = 298 K

At constant P and V,

Number of moles of gas (n) 1/(Absolute temperature (T))

n 1/T

nT = constant

n1T1 = n2T2 = …. = constant

n2T2 = nT1

n2 x 298 = 4.92 x 1010 x 303

n2 = (4.92 x 1010 x303)/298

n2 = 5.0025 x 1010

Number of N2 molecules to be added = n2 – n1= 5.0025 x 1010 - 4.92 x 1010

  Number of N2 molecules to be added =8.25 x 108 molecules.

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3)

Given data:

Molar mass of Helium (M) = 4 g (let’s kepp M in g so we will get density in g/L unit)

Pressure of the gas (P) = 1.07 x 10-6 torr = 1.07 x 10-6 x 0.0013 atm ……….( 1 torr = 0.0013 torr)

P = 1.391 x 10-9 atm

T = -252 oC = -252 + 273 = 21 K

R = 0.082 L.atm.K-1.mol-1.

Ideal gas equation for noble gas He,

PV = nRT

PV = (m/M) RT ………… (no. of moles = mass/molar mass i.e. m/Ma)

PM = (m/V) RT ……….. (rearranging equation)

m/V = mass / volume = density()

Let put it in equation,

PM = RT

= PM/RT

Let us put all known things,

= (1.391 x 10-9 x 4)/(0.082 x 21)

= 3.23 x 10-9 g/L

= 3.23 x 10-9 x g/1000mL

= 3.23 x 10-12 g/mL

Density of He gas at -252 oC and 1.07x10-6 torr pressure will be 3.23 x 10-12 g/mL.

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