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Based on a pressure due to neon of 0.8 atm, how many grams of neon are present if the temperature is 300 K, assuming it is contained in

a 2.5 L flask.
Use three significant figures.

1 Answer

3 votes

Answer:

1.34 grams of neon are present in the 2.5 L flask at a temperature of 300 K and a pressure of 0.8 atm.

Step-by-step explanation:

To determine the number of grams of neon present in a 2.5 L flask at a temperature of 300 K and a pressure of 0.8 atm, we can use the ideal gas law, which states that the pressure, volume, and temperature of a gas are directly proportional to the number of moles of the gas present.

We can rearrange the ideal gas law to solve for the number of moles of a gas present:

PV = nRT

Where:

P = the pressure of the gas

V = the volume of the gas

n = the number of moles of the gas

R = the universal gas constant

T = the temperature of the gas

In this case, we know the pressure, volume, and temperature of the gas, so we can solve for the number of moles of neon present:

n = PV / RT

Substituting in the values from the problem, we get:

n = (0.8 atm * 2.5 L) / (8.31 L * atm / mol * K * 300 K) = 0.067 mol

Since neon has a molar mass of 20.18 g/mol, we can convert the number of moles of neon to grams by multiplying the number of moles by the molar mass:

0.067 mol * 20.18 g/mol = 1.34 g

Therefore, there are approximately 1.34 grams of neon present in the 2.5 L flask at a temperature of 300 K and a pressure of 0.8 atm.

User Neal Davis
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