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2 years ago

Find the final pressure of a sample of gas at 1.3 atm and 1.9 l that is compressed to 158 ml

Chemistry

2 answers:

Alex Ar [27]2 years ago

7 0

Answer:

The final pressure of the gas is 15.6 atm (≅ 16 atm)

Explanation:

Given:

Initial volume of the gas, V1 = 1.9 L

Initial pressure, P1 = 1.3 atm

Final volume of the gas, V2 = 158 ml = 0.158 L

To determine:

Final pressure, P2

Explanation:

Based on the ideal gas equation

$PV = nRT$

where P = pressure, V = volume ; n = moles of gas

R = gas constant, T = temperature

At constant n and T, the above equation becomes:

PV = constant

This is the Boyle's law

Therefore:

$P1V1 = P2V2\\\\P2 = \frac{P1V1}{V2} = \frac{1.3 atm*1.9 L}{0.158L} =15.6 atm$

Fantom [35]2 years ago

6 0

From the Boyle's law of gases; the volume of a fixed mass is inversely proportional to the pressure at constant temperature;
Therefore; PV is a constant
Thus; P1V1 =P2V2
1.3 × 1900 ml = P2 × 158 ml
P2 = (1.3 × 1900) ÷ 158 ml
= 15.63 atm

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Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

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T = 20 + 273 = 293 K

Let's plug in the values in the equation:

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Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

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From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

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Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

0.195 = 0.453 - 2X

2X = 0.453 - 0.195

2X = 0.258

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So, there are 0.129 moles of methane in the mixture.

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