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marusya05 [52]
3 years ago
10

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

uation PV = C, where C is a constant. Suppose that at a certain instant the volume is 600 cm3, the pressure is 150 kPa, and the pressure is increasing at a rate of 40 kPa/min. At what rate is the volume decreasing at this instant?
Physics
1 answer:
gayaneshka [121]3 years ago
3 0

Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law,  PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}

Given;

\frac{dP}{dt} (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) =  600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( \frac{dv}{dt});

(600 x 40) + (150 x \frac{dv}{dt}) = 0

\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min

Therefore, the volume is decreasing at 160 cm³/min

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Answer:

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Explanation:

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The volume of sap in the vessel of length x is

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taking the derivative of both sides we get:

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A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2;

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2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v

solving for v we get:

v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}

\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}

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<h2>Greetings!</h2>

To find this value, you need to remember the speed formula:

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