
Why is warm air lighter than cold air?
(A) Due to heating, the pressure of air increases and air with higher pressure is lighter than air with lower pressure.
(B) Due to heating, the pressure of air decreases and air with lower pressure is lighter than air with higher pressure.
(C) Due to heating, air expands when the same mass occupies more volume it becomes lighter.
(D) Due to heating, air contracts when the same mass occupies less volume it becomes lighter.
Answer
221.4k+ views
Hint: For gases as temperature increases volume of gas increases i.e. if two gases have the same mass then gas at higher temperatures will have greater volume. So we can also say that the density of gas at higher temperatures will have a lesser density than the gas at lower temperatures.
Formula used
$\rho = \dfrac{M}{V}$ where$\rho $ is the density, $M$ is the mass, $V$ is the volume.
$V \propto T$ where $V$ is the volume, $T$is the temperature.
$P \propto T$ where $P$ is the pressure and $T$is the temperature.
Complete step by step solution:
We know that
$V \propto T$ where $V$ is the volume, $T$is the temperature. This only applies to the isobaric environment (constant pressure condition) and constant mass.
This means that for gases as the temperature increases the volume of gas increases i.e. if two gases have the same mass then gas at a higher temperature will have greater volume.
We know that
$\rho = \dfrac{M}{V}$ where$\rho $ is the density, $M$ is the mass, $V$ is the volume.
So we can also say that density of gas at a higher temperature (i.e. greater volume will have lesser density than the gas at a lower temperature.) and thus being lighter.
Hence we can assert that (C) Due to heating, air expands when the same mass occupies more volume it becomes lighter is the correct answer.
Additional information
Option (A) and (B) are incorrect as
$P \propto T$ ( where $P$ is the pressure and $T$ is the temperature ) this formula is only for the isochoric environment (constant volume condition) it could not be used for explanation purpose as in our atmosphere there is a natural tendency to maintain an isobaric environment (constant pressure condition).
Note:
Always remember the conditions for which a certain law is valid as in the question made by twist and turn of conditions are asked often and if you don’t remember them, then it can lead to deduction of your marks. The Isochoric environment is not valid in our nature.
Formula used
$\rho = \dfrac{M}{V}$ where$\rho $ is the density, $M$ is the mass, $V$ is the volume.
$V \propto T$ where $V$ is the volume, $T$is the temperature.
$P \propto T$ where $P$ is the pressure and $T$is the temperature.
Complete step by step solution:
We know that
$V \propto T$ where $V$ is the volume, $T$is the temperature. This only applies to the isobaric environment (constant pressure condition) and constant mass.
This means that for gases as the temperature increases the volume of gas increases i.e. if two gases have the same mass then gas at a higher temperature will have greater volume.
We know that
$\rho = \dfrac{M}{V}$ where$\rho $ is the density, $M$ is the mass, $V$ is the volume.
So we can also say that density of gas at a higher temperature (i.e. greater volume will have lesser density than the gas at a lower temperature.) and thus being lighter.
Hence we can assert that (C) Due to heating, air expands when the same mass occupies more volume it becomes lighter is the correct answer.
Additional information
Option (A) and (B) are incorrect as
$P \propto T$ ( where $P$ is the pressure and $T$ is the temperature ) this formula is only for the isochoric environment (constant volume condition) it could not be used for explanation purpose as in our atmosphere there is a natural tendency to maintain an isobaric environment (constant pressure condition).
Note:
Always remember the conditions for which a certain law is valid as in the question made by twist and turn of conditions are asked often and if you don’t remember them, then it can lead to deduction of your marks. The Isochoric environment is not valid in our nature.
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