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Arlecino [84]
3 years ago
6

Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f

ollowing?Part A:Decrease the volume to one-fourth the original volume while holding the temperature constant.Express your answer in terms of the variable P initial.Part B:Reduce the Kelvin temperature to half its original value while holding the volume constant.Express your answer in terms of the variable P initial .Part C:Reduce the amount of gas to half while keeping the volume and temperature constant.Express your answer in terms of the variable P initial .
Chemistry
1 answer:
pishuonlain [190]3 years ago
3 0

Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>

<em />

<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>

T₂ = 0.5 T₁. According to Gay-Lussac's law,

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}

P₂ in terms of the ideal gas equation is:

P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}

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Now, we have a total of 5 atoms of O on the left and 6 atoms on the right side. We can balance it by putting 3 in front of Fe2O3 and 6 in front of FeO as shown below:

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22. A piece of wood board contains approximately 6.76 X 1023 molecules of wood, C2H4920, how
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Answer:

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Explanation:

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From the question we have

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the first line

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