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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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Explanation:

Given that,

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Put the value of \omega=0.628\ rad/s in equation (I) and (II)

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tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

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Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

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Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

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tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

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From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

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