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Anna35 [415]
3 years ago
10

What pressure would be required to take 100.0 ml of a gas at 103.0 kPa and squish it down to 2.00 ml

Chemistry
1 answer:
melamori03 [73]3 years ago
7 0

Answer:

5150 kPa

Explanation:

Given that,

Initial volume, V₁ = 100 mL

Initial pressure, P₁ = 103 kPa

Final volume, V₂ = 2 mL

We need to find the new pressure of the gas. The relation between the pressure and the volume of gas is given by :

P_1V_1=P_2V_2\\\\P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{103\times 100}{2}\\\\P_2=5150\ kPa

So, the new pressure is 5150 kPa.

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The volume of a gas at 17.5 psi decreases from 1.8L to 750mL. What is the new pressure of the gas in arm?
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Answer:

P₂ = 2.88 atm

Explanation:

Given data:

Initial volume of gas = 1.8 L

Final volume = 750 mL

Initial pressure = 17.5 Psi

Final pressure = ?

Solution:

We will convert the units first:

Initial pressure = 17.5  /14.696 = 1.2 atm

Final volume = 750 mL ×1L/1000L = 0.75 L

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

1.2 atm × 1.8 L =  P₂  ×0.75 L

P₂ = 2.16 atm. L/ 0.75 L

P₂ = 2.88 atm

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