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4 years ago

5

At a constant temperature, the volume of a gas doubles when the pressure is reduced to half of its original value. This is a sta

tement of which gas law?

2 answers:

jolli1 [7]4 years ago

4 0

Answer:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

(At constant temperature and number of moles)

As pressure is decreased to half, the volume is increased to doubled.

Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

(At constant pressure and number of moles)

Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

(At constant volume and number of moles)

Combined gas Law: combining the three laws:

Explanation:

SIZIF [17.4K]4 years ago

3 0

<span>This is a statement of Boyle's Law.</span>

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HACTEHA [7]

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W = 135 J

The work done is 135 J.

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3 years ago

I need help on ideas for a science project.. (highschool ideas)

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2 years ago

How do I do number 10? Please show work!!

mafiozo [28]

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3 years ago

Does heat rise or fall?? the last person that answered did it just for the points and gave me no real answer

nasty-shy [4]

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3 years ago

Read 2 more answers

A composite load consists of three loads connected in parallel. One draws 100 W at a PF of 0.92 lagging, another takes 250 W at

fenix001 [56]

Answer:

a) $I_{RMS} = 4.79 A$

b) $PF = 0.908$

Explanation:

Get the reactive powers for each of the loads:

Reactive power = Real Power * tanθ

For load 1

Active power, P₁ = 100 W

Power factor, $cos \theta_{1} = 0.92$

$\theta_{1} = cos^{-1} 0.92\\\theta_{1} = 23.074$

$Q_{1}= P_{1} tan \theta_{1} \\Q_{1}= 100tan 23.074\\Q_{1}= 42.60 W$

For load 2

Active power, P₂ = 250 W

Power factor, $cos \theta_{2} = 0.8$

$\theta_{2} = cos^{-1} 0.8\\\theta_{2} = 36.87$

$Q_{2}= P_{1} tan \theta_{2} \\Q_{2}= 250tan 36.87\\Q_{2}= 187.5 W$

For load 3

Active power, P₃ = 250 W

Power factor, $cos \theta_{3} = 1$

$\theta_{3} = cos^{-1} 1\\\theta_{3} =0$

$Q_{2}= P_{1} tan \theta_{3} \\Q_{3}= 150tan 0\\Q_{3}= 0 W$

Calculate the total reactive power, $Q_{net} = 42.6 + 187.5 + 0$

$Q_{net} = 230.1 W$

Calculate the total active power, $P_{net} = 100 + 250 + 150 = 500 W$

$S_{net} = P_{net} + Q_{net} \\S_{net} = 500 + j230.1$

$P_{net} = IVcos \theta_{net}$

$\theta_{net} = tan^{-1} \frac{230.1}{500} \\\theta_{net} = 24.712$

V = 115 $V_{rms}$

$500 = I_{RMS} * 115 cos 24.712\\I_{RMS} = 500/104.47\\ I_{RMS} = 4.79 A$

b) Power factor of the composite load is $cos\theta_{net}$

$\theta_{net} = 24.712\\PF = cos 24.712\\PF = 0.908$

4 0

3 years ago