The pressure will increase 3 fold
It triples
If temperature and volume is fixed,pressure reduces.
The hydrogen in a sealed rubber party balloon is compressed slightly by the balloon rubber. So its density decreases a little, its pressure increases, and its temperature increases. But the temperature soon returns to the ambient temperature as heat is lost through the balloon wall. Also, the hydrogen will not stay in the balloon for long because it will leak out through pores in the rubber.
When an inflated balloon is exposed to cold air, provided pressure is constant, the volume will decrease. Bring the balloon back to a warmer spot, and the gas gains kinetic energy from the warm air, and the balloon will plump back up.
It doesn't matter as long as the conditions are the same. Temperature, gas pressure, speed with which you fill the balloon, and so on.
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The force of the pressure put on the balloon is squeezed into smaller particles so the volume is smaller. :) From Jade Nicole White. X
assuming the balloon is closed, the air pressure would double
If a balloon is squeezed, then that means the volume is decreasing. Volume and pressure vary indirectly, which means that when one goes up, the other goes down. So when you are decreasing the volume of the balloon, the pressure inside is going up (assuming constant mass and temperature).
If a balloon is squeezed, then that means the volume is decreasing. Volume and pressure vary indirectly, which means that when one goes up, the other goes down. So when you are decreasing the volume of the balloon, the pressure inside is going up (assuming constant mass and temperature).
Same throughout the ballon according to the Pascal's principle
Because hot air rises, the warmest air in the balloon is at the top of it. As time passes the balloonist has to fire the burners to maintain the temperature of the air in the balloon, and to prevent it being squeezed put by the higher pressure of the cold air surrounding the balloon.
Assuming the temperature is constant and none of the air is let out of the balloon, if you half the volume, the air pressure will double. If you want it, here is the math: Air is approximately an ideal gas, so when the temperature is constant, it follows the gas law of P1*V1 = P2*V2 where P1 is the pressure at state 1 (i.e. before the balloon is squeezed), V1 is the volume at state 1, P2 is the pressure at state 2 (i.e. after the balloon is squeezed), and V2 is the volume at state 2. When you half the volume, mathematically saying V2 = 0.5*V1. When you substitute this in, your equation now becomes P1*V1 = P2*0.5*V1. The V1 on both sides will cancel, leaving you with P1 = 0.5*P2. Since you are asking about the pressure after you squeeze the balloon, you want to know P2. So if you divide both sides by 0.5, you now have P1/0.5 = P2 or equivalently P2 = 2*P1 and we see that the pressure is twice that of the original.
The higher the temperature, the higher the pressure and vice versa. When the temperature rises the gas molecules move faster and hit the balloon more often and with more energy.
You could do this by applying pressure to the balloon, you'll notice that as it get's smaller from getting squeezed it becomes harder to compress, because of the higher pressure.
You could do this by applying pressure to the balloon, you'll notice that as it get's smaller from getting squeezed it becomes harder to compress, because of the higher pressure.
the pressure is 1.73289 ATM (atmospheres)
A weather balloon gathers information on temperature, atmospheric pressure, humidity, and wind speed.