
Density of air at room temperature is
(A)$120gc{m^{ - 3}}$
(B)$120gm{m^{ - 3}}$
(C)$12g{m^{ - 3}}$
(D)$1.3kg{m^{ - 3}}$
Answer
587.1k+ views
Hint:The measure of the closed package of the molecules are called the Density. The density of the air varies from one place with respect to another. The formula relates the temperature and the pressure. As the pressure increases the density also increases.
Complete step-by-step answer:
The density of air is given as mass per unit volume of the Earth’s atmosphere. That is:
$\rho = \dfrac{m}{v}$
Where $\rho $is the density of the unit of $kg{m^3}$, $m$is the mass and $v$ is the velocity.
Density of air at room temperature \[\left( {{{20}^o}C} \right)\]is \[1.2041kg{m^ - }^3\]
As we know air density is mass per unit volume of Earth’s atmosphere. It changes with change in atmospheric pressure, humidity and temperature. Therefore, air density is different at different temperatures.
Now if we look at the options nearest value to \[1.2041kg{m^ - }^3\]is \[1.3kg{m^{ - 3}}\]that is option (D).
Hence, option (D) is \[1.3kg{m^{ - 3}}\] correct option.
Additional information:
*Air density is used in many branches of science that deals with compressed air.
*Standard temperature and pressure are \[{100^o}C\]and \[100kPa\]. At this temperature and pressure air density is \[1.2754kg{m^{ - 3}}.\]
*Normal/room temperature and pressure is \[{20^o}C\]and \[101.325kPa\]. At this temperature and pressure air density is \[1.2041kg{m^ - }^3\].
*Air density changes with temperature because as temperature increases volume also increases because air molecules gain kinetic energy and move further apart. Therefore, as volume increases density decreases because volume is in the denominator.
*Similarly, as temperature decreases the kinetic energy of air molecules decreases as a result they come closer and volume decreases. Therefore, because volume is in the denominator, as volume decreases air density increases.
Note:
*Air density (air pressure) increases with increase in altitude.
*When all other parameters are the same, hot air will be less dense than cool air as a result it rises above the cool air.
Complete step-by-step answer:
The density of air is given as mass per unit volume of the Earth’s atmosphere. That is:
$\rho = \dfrac{m}{v}$
Where $\rho $is the density of the unit of $kg{m^3}$, $m$is the mass and $v$ is the velocity.
Density of air at room temperature \[\left( {{{20}^o}C} \right)\]is \[1.2041kg{m^ - }^3\]
As we know air density is mass per unit volume of Earth’s atmosphere. It changes with change in atmospheric pressure, humidity and temperature. Therefore, air density is different at different temperatures.
Now if we look at the options nearest value to \[1.2041kg{m^ - }^3\]is \[1.3kg{m^{ - 3}}\]that is option (D).
Hence, option (D) is \[1.3kg{m^{ - 3}}\] correct option.
Additional information:
*Air density is used in many branches of science that deals with compressed air.
*Standard temperature and pressure are \[{100^o}C\]and \[100kPa\]. At this temperature and pressure air density is \[1.2754kg{m^{ - 3}}.\]
*Normal/room temperature and pressure is \[{20^o}C\]and \[101.325kPa\]. At this temperature and pressure air density is \[1.2041kg{m^ - }^3\].
*Air density changes with temperature because as temperature increases volume also increases because air molecules gain kinetic energy and move further apart. Therefore, as volume increases density decreases because volume is in the denominator.
*Similarly, as temperature decreases the kinetic energy of air molecules decreases as a result they come closer and volume decreases. Therefore, because volume is in the denominator, as volume decreases air density increases.
Note:
*Air density (air pressure) increases with increase in altitude.
*When all other parameters are the same, hot air will be less dense than cool air as a result it rises above the cool air.
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