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tatiyna
3 years ago
15

The planet Jupiter has a surface temperature of 140 K and a mass 318 times that of Earth. Mercury (the planet) has a surface tem

perature between 600 K and 700 K and a mass 0.05 times that of Earth. On which planet is the atmosphere more likely to obey the ideal-gas law? Explain.
Physics
1 answer:
shtirl [24]3 years ago
6 0

Answer:

The atmosphere of Mercury likely to obey the ideal-gas law.

Explanation:

Every gas is an ideal gas when there are high temperatures and low pressure. In these conditions, the distance between the molecules is high, and inter-molecular interactions are minimum. According to the information, Mercury has a high temperature compared to Jupiter. That's why Mercury likely to obey the law of ideal-gas theory because there will be fewer interactions between the molecules.

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Assume a current of 1 ampere enters a parallel circuit at Point A. This 1 ampere of current will ________________ between Resist
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If a current of 1 ampere enters a parallel circuit at Point A. This 1 ampere of current will divide between Resistors R1 and R2 and then recombine at Point B

<h3>Parallel circuit</h3>

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Does spilled milk have higher or lower entropy
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What color is reflected off of most plant leaves?
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Green is reflected off of most plant leaves.
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The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
vovikov84 [41]

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Ratio, \frac{F_{1}}{F_{2}}= \frac{\frac{Gm_{1}M}{R^{2}}}{\frac{G3m_{1}M}{(3R)^{2}}}

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A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a
chubhunter [2.5K]

Answer:

The ballon will brust at

<em>Pmax = 518 Torr ≈ 0.687 Atm </em>

<em />

<em />

Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}

Rmax = 10.001 cm

Therefore, the max volume will be:

V_{max} = \frac{4}{3} \pi R_{max}^3

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Now, from (*)

P_{max} = P_1 \frac{V_1T_2}{V_2T_1}

Therefore:

Pmax= P1 * (0.687)

That is:

Pmax = 518 Torr

6 0
3 years ago
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