If a balloon is heated, the temperature inside the balloon increases, causing the air molecules to move faster and collide more frequently with the walls of the balloon. This increase in collisions leads to an increase in pressure inside the balloon, assuming the volume remains constant according to the ideal gas law (P1/T1 = P2/T2).
When a balloon is squeezed to half its volume at constant temperature, the air pressure inside the balloon increases. This is because the number of air molecules remains constant while the volume decreases, leading to the molecules being packed closer together and increasing the pressure.
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.
As a balloon rises, the pressure inside the balloon decreases. This is because the atmospheric pressure outside the balloon decreases with altitude, causing the balloon to expand as the pressure inside remains relatively constant.
Pressure increases.Take a look at the relationship:PV=nRTR is a constant. n is the amount of gas, which would be held constant in a sealed balloon. So:P1 V1 / T1 = P2 V2 / T21- before2- afterThis algebraic equation can help predict the outcome of other cases as well.taking just the balloon first, it is a solid body(flexible), on heating it undergoes cubical expansion, so the first thng that happens is the container(balloon) expands so untill a particular expansion of it the volume of air inside increases(gases dont have a fixed volume) they occupy the area they have .
For a balloon that is sealed and not full the volume of air inside the balloon will increase as it is heated. This is not however how hot air balloons work. A hot air balloon is essentially a fixed volume when it is inflated. If the air inside the balloon is heated the air inside becomes less dense so some of the air exits the balloon via the mouth of the balloon. As the air inside the balloon cools it becomes more dense so some air is ingested via the mouth of the balloon to keep it full. With each heating and cooling cycle, the pressure inside the balloon remains constant, the volume of the balloon remains constant but there is this movement of air out of and back into the balloon. P=VT Poop
When a balloon is squeezed to half its volume at constant temperature, the air pressure inside the balloon increases. This is because the number of air molecules remains constant while the volume decreases, leading to the molecules being packed closer together and increasing the pressure.
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.
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.
As a balloon rises, the pressure inside the balloon decreases. This is because the atmospheric pressure outside the balloon decreases with altitude, causing the balloon to expand as the pressure inside remains relatively constant.
The volume will increase in proportion to the increase in absolute temperature.
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.
Pressure increases.Take a look at the relationship:PV=nRTR is a constant. n is the amount of gas, which would be held constant in a sealed balloon. So:P1 V1 / T1 = P2 V2 / T21- before2- afterThis algebraic equation can help predict the outcome of other cases as well.taking just the balloon first, it is a solid body(flexible), on heating it undergoes cubical expansion, so the first thng that happens is the container(balloon) expands so untill a particular expansion of it the volume of air inside increases(gases dont have a fixed volume) they occupy the area they have .
For a balloon that is sealed and not full the volume of air inside the balloon will increase as it is heated. This is not however how hot air balloons work. A hot air balloon is essentially a fixed volume when it is inflated. If the air inside the balloon is heated the air inside becomes less dense so some of the air exits the balloon via the mouth of the balloon. As the air inside the balloon cools it becomes more dense so some air is ingested via the mouth of the balloon to keep it full. With each heating and cooling cycle, the pressure inside the balloon remains constant, the volume of the balloon remains constant but there is this movement of air out of and back into the balloon. P=VT Poop
If temperature remains constant and the volume of gas increases, the pressure will decrease. This is described by Boyle's Law, which states that pressure and volume are inversely proportional when temperature is constant.
remains constant
remains constant
The volume will increase in proportion to the increase in absolute temperature.