How Does Changing Gas Quantity and Temperature Affect Pressure?

In summary: Yes, the buoyant force is equal to the weight of the object floating. So in this case, the buoyant force must be equal to the weight of the blimp (helium included). To find the weight of the blimp, you need to use the equation F = mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s^2). So for part b, the answer would be "The buoyant force is equal to the weight of the blimp."For part c, you need to use the same equation (F = mg) but this time, m represents the maximum additional weight the blimp can carry while still maintaining a constant altitude. You can set up an
  • #1
Alethia
35
0
Hi, I'm having some trouble doing my homework for Physics. I would appreciate it if anybody would help me or lead me in the right direction. Thank you.

1. Gas is confined in a tank at a pressure of 1.0x18^8 Pa and a temperature of 15.0 degrees Celcius. If half the gas is withdrawm and the temperature is raised to 65.0 degrees Celcius, what is the new pressure in the tank in Pa?
---Normally for problems such as these, I use the prportion of (P1V1)/T1 = (P2V2)/T2. But I don't know how to solve without the volume. Is there a way I couild find the volume, or do I have to use a different forumla?

2. A 5450-m^3 blimp circles Fenway Park duing the World Series, suspended in the Earth's 1.21-kg/m^3 atmosphere. The density of the helium in the blimp is 0.178 kg/m^3. A) What is the buoyant force that suspends the blimp in the air? B) How does this buotant force compare to the blimp's weight? c) How much weight, in addition to the helium, can the blimp carry and still continue to maintain a constant altitude?
---I solved for a and for the answer 9.5x10^3 N by using the boyant force formula = D*g*v. However, I don't know what to do on b and c.


3. Allegra's favorite ride at the Barrel-O-Fun Amusement Park is the Flying Umbrella, which is lifted by a hydraulic jack. The operator activates the ride by applying a forcce of 72 N to a 3.0-cm-wide cylindrical piston, which holds the 20000. N ride off the ground. What is the diameter of the piston that holds the ride?
---Well honestly, for this problem, I have no idea where to start or do. n_n


Thank you! Anything helps and is appreciated.
 
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  • #2
Re part 1 of your question

Since the tank is presumably rigid, its volume is constant. So can't you just use P1/(N1T1) = P2/(N2T2) where N2=1/2 N1? And of course, don't forget to convert to absolute temperature when doing the ratio.

So P2 = (1/2) P1T2/T1.
 
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  • #3
Re part 3

The essential idea is that pressure is constant throughout the hydraulics. So you have P = F1/A1 = F2/A2. So A2=A1F2/F1. Let A1=3 cm, F1=72 N, F2=20,000 N. Then it's just D2=2 square root (A2/pi).
 
  • #4
Originally posted by Alethia
1. ...
---Normally for problems such as these, I use the prportion of (P1V1)/T1 = (P2V2)/T2. But I don't know how to solve without the volume. Is there a way I couild find the volume, or do I have to use a different forumla?
As Janitor points out, the volume is constant. But you can't just use P1/T1 = P2/T2. That's only true if the amount of gas (number of molecules) was fixed, which is not the case here. Use the entire ideal gas law, putting everything that's constant on one side. Then you can set up the right proportions.
2. ...
---I solved for a and for the answer 9.5x10^3 N by using the boyant force formula = D*g*v. However, I don't know what to do on b and c.
For b, if the thing is floating it must be in equilibrium. What does that mean? For c, realize that the bouyant force must be greater that just the weight of the helium---otherwise what's the point? You wouldn't be able to lift anything but the helium.
3. ...
---Well honestly, for this problem, I have no idea where to start or do.
For hydraulic jacks, realize that the pressure in the fluid is the same on both sides. Find the pressure on one side, then use that information to figure out how wide the other piston must be. (Set it up as a ratio.)
 
  • #5
Okay for number 1 I calculated it, and I ended up with 5.7x10^7 Pa. Is that right?

Uhm, if it is in equilibrium it is at zero. So does that mean that the boyant force cancels with the blimp's weight? And for c, yeah, it would make sense that the buoyant force is larger... but how would I calculate it?

On the last problem, why is A1 = 3 cm? And how did you conclude it to this: D2=2 square root (A2/pi)? I'm sorry, but could you break it down a little more?

Thank you both! Much appreciated!
 
  • #6
My goof.

When I was writing up some hints on Q 3, I was thinking the area of the cylinder bore was 3 square centimeters, but I see that in fact the diameter was 3 cm. My mistake, sorry. Anyway, since A = pi r^2, then A = pi (D/2)^2. Inverting this, D = 2 square root (A/pi).
 
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  • #7
As far as the numerical answer to Q 1 goes, did you add 273 to each Celsius temperature to get it into Kelvins? If so, then you are probably okay. I don't have a calculator handy. When you wrote " 1.0x18^8 Pa" did you mean 1.0x10^8 Pa?
 
  • #8
Originally posted by Janitor
As far as the numerical answer to Q 1 goes, did you add 273 to each Celsius temperature to get it into Kelvins? If so, then you are probably okay. I don't have a calculator handy. When you wrote " 1.0x18^8 Pa" did you mean 1.0x10^8 Pa?
Yeah I meant x10^8. Whoopos. =) And yes, I did convert everything to Kelvins.
 
  • #9


Originally posted by Janitor
When I was writing up some hints on Q 3, I was thinking the area of the cylinder bore was 3 square centimeters, but I see that in fact the diameter was 3 cm. My mistake, sorry. Anyway, since A = pi r^2, then A = pi (D/2)^2. Inverting this, D = 2 square root (A/pi).
Oh! Thank you! I calculated it and it came out as 50 cm even. =) Thanks!
 
  • #10
Originally posted by Alethia
Uhm, if it is in equilibrium it is at zero. So does that mean that the boyant force cancels with the blimp's weight? And for c, yeah, it would make sense that the buoyant force is larger... but how would I calculate it?
For B: Yes, the bouyant force equals the total weight of the blimp.
For C: Calculate the weight of the helium (using density and volume) and subtract that from the total weight (which equals the bouyant force).
 
  • #11
1. Gas is confined in a tank at a pressure of 1.0x18^8 Pa and a temperature of 15.0 degrees Celcius. If half the gas is withdrawm and the temperature is raised to 65.0 degrees Celcius, what is the new pressure in the tank in Pa?

Withdrawing half of the gas will half the pressure. Raising the temperature by ? percent (figure it out yourself, in kelvins) will raise the pressure by the same percent.
 

1. What is pressure?

Pressure is defined as the force applied per unit area. In other words, it is the amount of force exerted on a surface divided by the area of that surface.

2. How is pressure measured?

Pressure is typically measured using a device called a manometer, which measures the difference in height of a liquid in a column. The most common unit of pressure is the Pascal (Pa), but other units such as atmospheres (atm) and millimeters of mercury (mmHg) are also commonly used.

3. What is the Ideal Gas Law?

The Ideal Gas Law is a mathematical equation that relates the pressure, volume, temperature, and number of moles of an ideal gas. It is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

4. What is an ideal gas?

An ideal gas is a theoretical gas that follows the Ideal Gas Law under all conditions. In other words, it has no intermolecular forces, occupies no volume, and has no attractive or repulsive forces between molecules. Real gases behave similarly to ideal gases under certain conditions, but deviate at high pressures and low temperatures.

5. How does temperature affect pressure?

According to the Ideal Gas Law, temperature and pressure are directly proportional when volume and number of moles are constant. This means that as temperature increases, so does pressure, and vice versa. This relationship is known as Charles' Law and is one of the fundamental principles of the Ideal Gas Law.

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