Pressure and temperature affect the amount of space between gas molecules, which

Question

Pressure and temperature affect the amount of space between gas molecules, which affects the volume and, therefore, the density of the gas since density volume The molar mass of a substance, however, is a constant and can be used to identify an unknown gas sample. density = mass/volume The Molar mass is found by dividing the mass of a sample (in grams) by the number of moles in that sample. The number of moles of gas can be calculated using the ideal gas law PV = nRT which can be rearranged as n = PV/RT Given the number of moles of a gas and its molar mass, you can calculate the mass of the gas Since density is equal to the ratio of the mass and volume, you can then divide by the volume to find density Alternatively, you can use the ratio n/V the ideal gas equation where n is the number of moles and Vis the volume, and convert from moles per unit volume to grams per unit volume using molar. Calculate the density of oxygen, O_2, under each of the following conditions. STP 1,00 atm and 30.0 degree C. Express your answers numerically in grams per liter. Enter the density at STP first and separate answers by a comma. density at STP, density at 1 atm and 30.0 degree C = To identify a diatomic gas (X_2) a researcher carried out the following experiment She N weighed an empty 3.5-L bulb, then filled it with the gas at 1.60 atm and 27.0 degree C and weighed it again. The difference in mass was 6.3 g. Identify the gas. Express your answer as a chemical formula.

Explanation / Answer

Using gas equation,

d =PM/RT

where,

d= density

P=Pressure

M=Molecular mass

T= Temperature

Now,

At STP,

Temperature=273 K

And, Pressure = 1 atm

Putting these values in Gas equation, we get,

d =PM/RT

=(1 atm x 32 g mol-1 )/0.0821 L atm K -1mol-1 x 273 K

=1.4277 g L-1

Similarly,

Density at 1.00 atm and 300 C (303 K)=

d =PM/RT

=(1 atm x 32 g mol-1)/ 0.0821 L atm K -1mol-1 x 303 K

=1.2863 g L-1

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