
Arrange the following in order of increasing density?
Oxygen at $25^\circ C$ , $1{\text{atm}}$ ; Oxygen at $0^\circ C$ , $2{\text{atm}}$ ; Oxygen at $273^\circ C$ , $1{\text{atm}}$ .
Answer
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Hint: We have to know that the thickness can be changed by changing either the pressing factor or the temperature. Expanding the pressing factor consistently builds the thickness of a material. Expanding the temperature, for the most part, diminishes the thickness, yet there are striking exemptions for this speculation.
Complete answer:
We have to know that the impact of pressing factor and temperature on the densities of fluids and solids is little. The compressibility for a regular fluid or strong is ${10^{ - 6}}ba{r^{ - 1}}$ and a run of the mill warm expansivity is ${10^{ - 5}}{K^{ - 1}}$ . This generally converts into requiring around multiple times environmental strain to diminish the volume of a substance by one percent. (Albeit the pressing factors required are possibly around multiple times more modest for sandy soil and a few muds.) A one percent extension of volume ordinarily requires a temperature increment on the request for a huge number of degrees Celsius.
Conversely, the thickness of gases is unequivocally, influenced by pressure. The thickness of an ideal gas is,
$d = \dfrac{{MP}}{{RT}}$
Where,
$M$ is the molar mass,
$P$ is the pressing factor,
$R$ is the general gas constant,
$T$ is the outright temperature.
This implies that the thickness of an ideal gas can be multiplied by multiplying the pressing factor, or by splitting the total temperature.
For, oxygen at $25^\circ C$ = $298K$, $1{\text{atm}}$ ;
$\dfrac{P}{T} = \dfrac{1}{{298}}$
For, oxygen at $0^\circ C$ = $273K$ , $2{\text{atm}}$ ;
$\dfrac{P}{T} = \dfrac{2}{{273}}$
For, oxygen at $273^\circ C$ = $546K$ , $1{\text{atm}}$ ;
$\dfrac{P}{T} = \dfrac{1}{{546}}$
Therefore,
The increasing order of density is,
Oxygen at $273^\circ C$ , $1{\text{atm}}$ , Oxygen at $25^\circ C$ , $1{\text{atm}}$ ; Oxygen at $0^\circ C$ , $2{\text{atm}}$ .
Note:
We have to know that, similarly for the decreasing order of density, that will be given below, Oxygen at $0^\circ C$ , $2{\text{atm}}$ ; Oxygen at $25^\circ C$ , $1{\text{atm}}$ ; Oxygen at $273^\circ C$ , $1{\text{atm}}$.
Complete answer:
We have to know that the impact of pressing factor and temperature on the densities of fluids and solids is little. The compressibility for a regular fluid or strong is ${10^{ - 6}}ba{r^{ - 1}}$ and a run of the mill warm expansivity is ${10^{ - 5}}{K^{ - 1}}$ . This generally converts into requiring around multiple times environmental strain to diminish the volume of a substance by one percent. (Albeit the pressing factors required are possibly around multiple times more modest for sandy soil and a few muds.) A one percent extension of volume ordinarily requires a temperature increment on the request for a huge number of degrees Celsius.
Conversely, the thickness of gases is unequivocally, influenced by pressure. The thickness of an ideal gas is,
$d = \dfrac{{MP}}{{RT}}$
Where,
$M$ is the molar mass,
$P$ is the pressing factor,
$R$ is the general gas constant,
$T$ is the outright temperature.
This implies that the thickness of an ideal gas can be multiplied by multiplying the pressing factor, or by splitting the total temperature.
For, oxygen at $25^\circ C$ = $298K$, $1{\text{atm}}$ ;
$\dfrac{P}{T} = \dfrac{1}{{298}}$
For, oxygen at $0^\circ C$ = $273K$ , $2{\text{atm}}$ ;
$\dfrac{P}{T} = \dfrac{2}{{273}}$
For, oxygen at $273^\circ C$ = $546K$ , $1{\text{atm}}$ ;
$\dfrac{P}{T} = \dfrac{1}{{546}}$
Therefore,
The increasing order of density is,
Oxygen at $273^\circ C$ , $1{\text{atm}}$ , Oxygen at $25^\circ C$ , $1{\text{atm}}$ ; Oxygen at $0^\circ C$ , $2{\text{atm}}$ .
Note:
We have to know that, similarly for the decreasing order of density, that will be given below, Oxygen at $0^\circ C$ , $2{\text{atm}}$ ; Oxygen at $25^\circ C$ , $1{\text{atm}}$ ; Oxygen at $273^\circ C$ , $1{\text{atm}}$.
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