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Maxine
TaigaWhen a balloon is heated, the air inside the balloon expands and the molecules move faster, increasing their kinetic energy. This results in a decrease in mass density, as the molecules are more spread out. However, the total mass of the air inside the balloon remains the same.
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What happens To the mass of air in a balloon when it is heated?
When air inside a balloon is heated, the molecules move faster, causing the air to expand and the balloon to inflate. The total mass of the air inside the balloon remains the same, but the density of the air decreases as it becomes less compact.
What happens to the volume in a balloon when heated?
When a balloon is heated, the air molecules inside the balloon gain kinetic energy and move faster, causing them to spread out and take up more space. This increase in volume leads to the balloon expanding.
What happens to gas in a balloon when that gas is heated?
When the gas in a balloon is heated, its temperature increases, causing the gas molecules to move faster and collide more frequently with the walls of the balloon. This increases the pressure inside the balloon, which causes the balloon to expand as it attempts to accommodate the increased volume of gas.
When the air in a balloon is heated what happens to the density?
When the air inside a balloon is heated, its density decreases. This happens because the air molecules gain energy and move farther apart, resulting in a decrease in the air's density.
What happens to the mass of a balloon when you put it in a warm room?
Nothing happens to the mass of the balloon. Mass is conserved, so the temperature of the balloon will not affect it's mass. Mass can be thought of the amount of "stuff" that makes up a balloon. It can be obtained by adding up the mass of all the molecules of rubber in the balloon. Obviously, putting the balloon in a warm room will not change the number of molecules in the balloon, therefore the mass stays constant. The volume of the balloon will probably increase. Because volume increases but mass remains constant, the density of the balloon would decrease. D = m/v
