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PolarNik [594]
3 years ago
7

A piston is filled with gas. When the pressure is increasing, what is happening to the volume? Assume that all other properties

of the gas remain constant.
a. It is remaining the same.
b. It is increasing.
c. It is decreasing.
d. It is increasing or decreasing until it reaches zero.
Physics
1 answer:
Kryger [21]3 years ago
3 0
The correct answer is letter C. Volume is decreasing. For a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume<span> are </span>inversely proportional<span>. </span>
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According to Georgia State University, gravitational potential energy<span> is the energy an object possesses because of its position in a gravitational field. This is most commonly in reference to an object near the surface of the Earth, where the gravitational acceleration is assumed to be constant at about 9.8 m/s2.</span>
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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
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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:

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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

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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} }

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

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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

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