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Shkiper50 [21]
2 years ago
5

Students want to investigate the inverse relationship between the pressure and temperature of an ideal gas as predicted by the i

deal gas law. Their plan is to use a gas filled cylinder with a movable piston on one end and a heater inside that can be turned on and off. The students will the measure the pressure and temperature of the gas. Which of the following refinements to this procedure will allow the students to observe the predicted relationship between pressure and temperature? Select two answers
A. Start with enough gas to have a pressure near atmospheric pressure, and repeat the experiment, removing gas from the cylinder each time.
B. Fix the piston in place so the volume of the pas remains constant.
C. Ensure the piston and cylinder walls are insulated to the gas can reach equilibrium for each set of measurements
D. Conduct the investigation under conditions of very high pressure to ensure ideal gas behavior
Physics
1 answer:
grin007 [14]2 years ago
7 0

Answer:

Option B, Fix the piston in place so the volume of the pas remains constant

Explanation:

As we know

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

The effect on variable due to another variable can be studied by keeping the third variable constant.

Hence, in order the study the variation of temperature with pressure or vice versa, the volume needs to fixed at a certain value.

Hence, option B is correct

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A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a
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Explanation:

Given that,

Mass if the rock, m = 1 kg

It is  suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

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5 0
3 years ago
While bats emit a wide variety of sounds, one type emits pulses of sound at a frequency between 39 kHz and 78 kHz. What is the r
sdas [7]

Answer:

The range of wavelengths of the sound is 7692.30 m and 3846.15 m

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If f = 39 kHz,

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{39\times 10^3}\\\\\lambda=7692.30\ m

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\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{78\times 10^3}\\\\\lambda=3846.15\ m

So, the range of wavelengths of the sound is 7692.30 m and 3846.15 m.

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3 years ago
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