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vitfil [10]
3 years ago
8

A rigid tank of air is cooled (temperature decreased). The volume has to remain constant because it is a rigid tank. The air pre

ssure in the rigid tank
A. Increases B. Decreases C. Stays the same
Physics
2 answers:
Romashka-Z-Leto [24]3 years ago
5 0

The air pressure in the rigid tank Decreases.

<u>Explanation:</u>

A rigid tank is said to be a region where volume is always considered as a constant. During the process of cooling the rigid tank where the temperature is said to be decreased, the effect of air pressure in the tank also decreases.

Because, PV = RT, where P is the pressure, T is temperature, R is constant, V is the volume. Hence pressure is directly proportional to temperature.

alekssr [168]3 years ago
4 0
If the temperature in the tank is decreased this would imply that the kinetic energy of molecule decreases because of the ideal gas law:
PV= nRT (Since Pressure is directly proportional to Temperature).
So the answer is B.The air pressure in the rigid tank decreases.
Hope this helps you.
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If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is
expeople1 [14]

Answer:

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v2^2 = 2 g H + v1^2

Since 88 ft/sec = 60mph   we have 30 mph = 44 ft/sec

The object will return with the same speed that it had initially so the object

starts out with a downward speed of 44 ft/sec

Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2

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8 0
3 years ago
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ipn [44]

Answer:

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Explanation:

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Substitute the given values

ΔU=970J-223J

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For(b) γ for the gas.

We can calculate γ by ratio of heat capacities of the gas

γ=Cp/Cv

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To calculate γ we first need to find Cp and Cv

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As we know

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Cp=(Q/nΔT)

C_{p}=\frac{970J}{1.75mol*(25^{o}C-10^{o}C )}\\C_{p}=37J/mol.K

From relation of Cv and Cp we know that

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So

C_{v}=C_{p}-R\\C_{v}=37-8.314\\C_{v}=28.687J/mol.K\\

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γ=[(37J/mol.K) / (28.687J/mol.K)]

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3 years ago
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Answer:

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Viefleur [7K]

Answer:

Explanation:

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