Which contribution did Kepler make to viewing the solar system?

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

3 years ago

In the four-stroke internal combustion engine, fuel is ignited during the ____ stroke.

forsale [732]

The four strokes in order are the intake stroke, the compression stroke, the power stroke, and the exhaust stroke. Fuel is ignited during the power stroke.

7 0

3 years ago

Read 2 more answers

A weightlifter liftsa 1,250-N barbell 2 m in 3 s. how much power was used to lift the barbell?

STatiana [176]

Power = Force * Distance/ time
P = 1,250 * 2/3
P = 2,500/3
P = 833.33 Watts

So, your final answer is 833.33 Watts

5 0

4 years ago

How does the law of conservation of energy apply to a light bulb? the electrical energy put into the light bulb equals the light

andrew-mc [135]

<span>The law of conservation of energy applies to a light bulb because the energy is being transformed into light and the light bulb is acting as a catalyst. The light bulb itself is not a form of energy, however when in combination with the electrical outlet to the bulb the electricity heats up the metal interior forming it into light. according to the law of conservation energy cannot be created or destroyed, but instead is formed into different kinds of energy. In relation to a light bulb electrical currents are forming heat energy by heating up the metal interior, then the bulb or glass around it allows to radiate light.</span>

7 0

3 years ago

A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin

Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

$\epsilon = \frac{NBA}{t}$