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If 1.9 g of oxygen gas (O2) occupies a volume of 100 L at a particular temperature and pressure, what volume will 5.00 g of oxygen gas occupy under the same conditions?

User Porglezomp
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To solve this problem, we can use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas.

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

Since we are dealing with the same gas (oxygen) at the same temperature and pressure, we can assume that the value of R is constant. Therefore, we can write:

PV/n = RT

The ratio PV/n is known as the molar volume of the gas, and is constant for a given temperature and pressure. We can use this relationship to find the volume of oxygen gas that corresponds to 5.00 g.

First, we need to calculate the number of moles of oxygen gas in 1.9 g:

n = m/M

where m is the mass of the gas and M is the molar mass of oxygen, which is 32 g/mol.

n = 1.9 g / 32 g/mol

n = 0.059375 mol

The molar volume of oxygen gas at the given temperature and pressure is:

PV/n = RT/n

V/n = RT/P

Substituting the given values, we get:

V/n = (0.0821 L·atm/mol·K) x (273 K) / (1 atm)

V/n = 22.414 L/mol

Therefore, the volume of 1.9 g of oxygen gas is:

V1 = n x V/n = 0.059375 mol x 22.414 L/mol = 1.331 L

Now we can use the molar volume to find the volume of 5.00 g of oxygen gas:

n = m/M = 5.00 g / 32 g/mol = 0.15625 mol

V2 = n x V/n = 0.15625 mol x 22.414 L/mol = 3.51 L

Therefore, 5.00 g of oxygen gas will occupy a volume of 3.51 L at the same temperature and pressure as 1.9 g of oxygen gas.

User Norman Xu
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