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pav-90 [236]
4 years ago
13

A gas at constant temperature occupies a volume of 2.40 L and exerts a pressure of 0.85 atm. What volume will the gas occupy at

a pressure of 0.15 atm? Work must be shown in order to earn credit.
Chemistry
1 answer:
sesenic [268]4 years ago
4 0

Answer:

13.6L

Explanation:

Given parameters:

Initial Volume V₁ = 2.4L

Initial pressure P₁ = 0.85atm

Final pressure P₂ = 0.15atm

Final volume V₂ = ?

Solution

To solve the problem we have to take the condition stated in the problem very carefully. It was stated that the gas was at a constant temperature.

According to Boyle's law" The volume of a fixed mass of a gas varies inversely as the pressure changes if the temperature is constant".

We can simply apply this law to solve the problem:

The law is mathematically expressed as P₁V₁ = P₂V₂

     

The unknown here is V₂ the final volume. We express it as the subject of the formula:

                     V₂ = \frac{P_{1} V_{1} }{P_{2} }

Now solving for the final volume, we have:

                 V₂ = \frac{2.4 x 0.85}{0.15}

                 V₂ = \frac{2.04}{0.15} = 13.6L

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A 7.00 L tank at 21.4^oC is filled with 5.43 g of sulfur hexafluoride gas and 14.2 g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas. Round each of your answers to significant digits.

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<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

  • <u>For sulfur hexafluoride:</u>

Given mass of sulfur hexafluoride = 5.43 g

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Putting values in equation 1, we get:

\text{Moles of sulfur hexafluoride}=\frac{5.43g}{146.06g/mol}=0.0372mol

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Given mass of sulfur tetrafluoride = 14.2 g

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Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

\chi_A=\frac{n_A}{n_A+n_B} .....(2)

where n is the number of moles

Putting values in equation 2, we get:

\chi_{SF_6}=\frac{0.0372}{0.1686}=0.221

\chi_{SF_4}=\frac{0.1314}{0.1686}=0.779

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