Question wasn\'t fully answered. A 2.95-L gas sample in a cylinder with a freely

Question

Question wasn't fully answered.

A 2.95-L gas sample in a cylinder with a freely moveable piston is initially at 33.0 ?C. What is the volume of the gas when it is heated to 141 ?C (at constant pressure)?

What is the change in temperature of a 2.50 L system when its volume is reduced to 1.60 L if the initial temperature was 298 K?

A gas sample occupies 3.50 liters of volume at 20.°C. What volume will this gas occupy at 100°C (reported to three significant figures)?

If you double the pressure of a constant amount of gas at a constant temperature, what happens to the volume?

What is the final volume (L) of a 10.0 L system that has the pressure tripled?

A balloon filled with 0.500 L of air at sea level is submerged in the water to a depth that produces a pressure of 3.25 atm. What is the volume of the balloon at this depth?

What is the equivalent pressure of 0.905 atm in units of mm Hg?

What is the equivalent pressure of 760 torr in units of mm Hg?

have an average kinetic energy proportional to the temperature in kelvinsare packed closely together?nearly touching one anotherattract one another strongly

all of the above

All of the following statements are consistent with the kinetic molecular theory of gases EXCEPT:

The size of the gas molecules is negligible compared to the total volume of the gas.The average kinetic energy of the molecules of a gas is proportional to the temperature of the gas in kelvins.Strong attractive forces hold the gas molecules together.The gas molecules collide with each other and with the surfaces around them.none of the above

Explanation / Answer

Solution:- (1) V1 = 2.95 L, T1 = 33.0 + 273 = 306 K, T2 = 141 + 273 = 414 K, V2 = ?

It's Charle's law as the pressure is constant and the volume is changing along with kelvin temperature.

V1T2 = V2T1

V2 = V1T2/T1

V2 = (2.95 L x 414 K)/306 K

V2 = 3.99 L

(2) V12.50 L, V2 = 1.60 L, T1 = 298 K, T2 = ?

It's Charle's law again as the pressure is constant.

V1T2 = V2T1

T2 = V2T1/V1

T2 = (1.60 L x 298 K)/2.50 L

T2 = 191 K

So, change in temperature = 191 K - 298 K = -107 K

(3) V1 = 3.50 L, T1 = 20+273 = 293K, T2 = 100+273 = 373K, V2 = ?

V2 = V1T2/T1

V2 = (3.50 L x 373K)/293K

V2 = 4.46 L

(4) From Boyel's law, "At constant temperature, volume is inversely proportional to the pressure."

P1V1 = P2V2

V2 = P1V1/P2

It says pressure is doubled that means P2 = 2P1

V2 = P1V1/2P1

V2 = V1/2

So, if the pressure is doubled then the volume is halved.

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