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4 years ago

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A gas has a pressure of 48atm in a 15.5L container. It was found that at 25∘C the gas occupied a volume of 25L and had a pressur

e of 22atm. What was the initial temperature in degrees Celsius?

1 answer:

EleoNora [17]4 years ago

3 0

Answer:

130.165636364°C

Explanation:

P = Pressure

V = Volume

n = Number of moles

R = Gas constant = 0.082 L atm/mol K

From ideal gas law we have

$PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{22\times 25}{0.082\times (25+273.15)}\\\Rightarrow n=22.496451696\ moles$

$PV=nRT\\\Rightarrow T=\dfrac{PV}{nR}\\\Rightarrow T=\dfrac{48\times 15.5}{22.496451696\times 0.082}\\\Rightarrow T=403.315636364\ K$

The initial temperature is $403.315636364-273.15=130.165636364\ ^{\circ}C$

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a) The launch velocity of the rocket is 5.48 m/s

b) The maximum height is 1.53 m

Explanation:

a)

We can solve this part by applying the law of conservation of energy, by considering the kinetic energy and the elastic potential energy only, since there is no change in gravitational potential energy and no friction is involved.

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x = 0.10 m (compression of the spring)

The total energy when the spring is relased is:

$E=KE_f + PE_{sf}$

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In this part instead we consider only the kinetic energy and the gravitational potential energy, since the spring is at rest so its energy is now zero.

The total energy at the launch is:

$E=KE_i + PE_{gi}$

where

$KE_i = \frac{1}{2}mv^2$ (initial kinetic energy), with

m = 0.15 kg (mass of the rocket)

v = 5.48 m/s (velocity of launch of the rocket)

$PE_{gi}=0$ (initial gravitational potential energy is zero)

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where

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h = ? (maximum height)

Combining the two equations we find

$\frac{1}{2}mv^2 = mgh$

And solving for h,

$h=\frac{v^2}{2g}=\frac{(5.48)^2}{2(9.8)}=1.53 m$

Learn more about potential energy and kinetic energy:

brainly.com/question/1198647

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brainly.com/question/6536722

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