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CaHeK987 [17]
1 year ago
7

A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 l to 15 l. the final pressure is ________ atm

. group of answer choices
Physics
1 answer:
LekaFEV [45]1 year ago
6 0

A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.

A physics principle states that a gas's volume changes inversely with applied pressure while the gas is kept at a constant temperature. Boyle's law states that in constant temperature the  variation volume of gas is inversely proportional to the applied pressure. According to Boyle's law, a gas's pressure is inversely proportional to its volume while the gas is at constant temperature.  To put it another way, we can argue that volume and pressure are inversely proportional to one another, but only for gases with constant mass and temperature.

The formula is,

P₁ X V₁ = P₂ X V₂

Where,

P₁ is initial pressure = 1 atm

P₂ is final pressure =  ? (Not Known)

V₁ is initial volume  = 10 L

V₂ is final volume = 15 L

Now put the values in the formula,

1 x 10 = P₂ x 15

10 = 15 P₂

P₂ = 0.67

Therefore, the final pressure is 0.67 atm.

Learn more about final pressure problem here:

brainly.com/question/14221674

#SPJ4

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Answer:

% reduction in area = 54.26 %

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Explanation:

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% reduction in area =\frac{(108.8 - 49.97)}{108.8} * 100 = 54.26 %

b)percentage elongation = \frac{(66 - 46.1)}{46.1} *100 = 43.16 %

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The freight cars a and b have a mass of 20 mg and 15 mg, respectively. if the cars collide and couple together, what is the velo
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Explain about kinetic theory​
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When flying in an airplane, you are most likely in which layer of the atmosphere? mesosphere thermosphere stratosphere trosphere
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A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg
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Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

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Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

280 = 5 v₁² + (10 − 5 v₁)² / 4

1120 = 20 v₁² + (10 − 5 v₁)²

1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²

0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

v₁ = -3.78 m/s or 6 m/s

u₁ = 6 m/s, so v₁ = -3.78 m/s.  Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

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