0
What else can I help you with?
What happens to the air pressure inside a balloon when the balloon is squeezed to half it volume at constant temperature?
The air pressure inside the balloon will increase when it is squeezed to half its volume at constant temperature. This is because the volume of the balloon decreases, leading to the air molecules being more confined in a smaller space, resulting in higher pressure.
What happens to the air pressure inside a balloon when the balloon is squeezed to half its volume at constant temperature?
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.
If a balloon is squeezed what happens to the gas inside?
When a balloon is squeezed, the volume of the balloon decreases. This causes the gas inside the balloon to be compressed, increasing the pressure of the gas.
If a balloon is rubbed vigorously what happens to the pressure of the air inside the balloon if the volume remains constant?
The pressure inside the balloon will increase due to the increase in temperature caused by the rubbing (which is a form of mechanical work). According to the ideal gas law, pressure is directly proportional to temperature when volume is constant.
What is the effect of forces exerted on a balloon when it is squeezed?
When a balloon is squeezed, the forces exerted on it cause the air inside the balloon to be compressed. This compression increases the pressure inside the balloon, leading to a change in the balloon's shape and size. If the squeezing force is too strong, it can cause the balloon to burst.
What happens to the air pressure inside a balloon when the balloon is squeezed to half it volume at constant temperature?
The air pressure inside the balloon will increase when it is squeezed to half its volume at constant temperature. This is because the volume of the balloon decreases, leading to the air molecules being more confined in a smaller space, resulting in higher pressure.
If a balloon is squeezed what a happens to the pressure of the gas inside the balloon?
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 what happens to the pressure of the gas inside the balloon?
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).
What happens to the air pressure inside a balloon when the balloon is squeezed to half its volume at constant temperature?
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.
What change in pressure occurs in a party balloon that is squeezed to one-third its volume with no change in temperature?
The pressure of the balloon increases threefold when it is squeezed to one-third its volume with no change in temperature. This is because pressure is inversely proportional to volume according to Boyle's Law, which states that when volume decreases, pressure increases.
If a balloon is squeezed what happens to the gas inside?
When a balloon is squeezed, the volume of the balloon decreases. This causes the gas inside the balloon to be compressed, increasing the pressure of the gas.
Is it possible for a balloon with an initial pressure of 200.0kPa to naturally expand four times its initial volume when the temperature remains constant and atmospheric pressure is 101.3kPa?
No, it is not possible for the balloon to naturally expand four times its initial volume while the temperature remains constant. According to Boyle's Law, at constant temperature, the pressure and volume of a gas are inversely proportional. Since the atmospheric pressure remains constant, the balloon's pressure of 200.0kPa would need to increase to expand, which cannot happen at constant temperature.
How is isothermal possible in a balloon?
Isothermal is where pressure and/or volume changes, but temperature remains constant. Pressure, Volume, and Temperature are related as: PV = nRT =NkT for an ideal gas. Here, we see that since a balloon's volume is allowed to change, its pressure remains relatively constant. Whenever there is a pressure change, it'll be offset by an equivalent change in volume, thus temperature is constant.
If a balloon is rubbed vigorously what happens to the pressure of the air inside the balloon if the volume remains constant?
The pressure inside the balloon will increase due to the increase in temperature caused by the rubbing (which is a form of mechanical work). According to the ideal gas law, pressure is directly proportional to temperature when volume is constant.
Suppose the temperature of the air in a balloon is increased if the pressure remains constant.. what quantity must change Would it be Volume or Number of molecules or Compressibility OR Adhesion?
If the pressure is kept constant while increasing the temperature of the air in a balloon, the volume of the gas inside the balloon would change. This is because as the temperature rises, the gas molecules gain energy and move faster, leading to an increase in volume to maintain a constant pressure.
How will volume of a balloon change if pressure remains constant but pressure increase?
The volume will increase in proportion to the increase in absolute temperature.
How will volume of a balloon change if pressure remains constant but temperature increases?
The volume will increase in proportion to the increase in absolute temperature.
