Respuesta :
Answer: [tex]R=4.82436 \frac{Pa. m^{3}}{mol. K}[/tex]
Explanation:
The Ideal Gas equation is:
[tex]P.V=n.R.T[/tex] (1)
Where:
[tex]P[/tex] is the pressure of the gas
[tex]n[/tex] the number of moles of gas
[tex]R=8.3144598 \frac{Pa. m^{3}}{mol. K}[/tex] is the gas constant
[tex]T[/tex] is the absolute temperature of the gas in Kelvin.
[tex]V[/tex] is the volume
It is important to note that the behavior of a real gas is far from that of an ideal gas, taking into account that an ideal gas is a single hypothetical gas. However, under specific conditions of standard temperature and pressure (T=0\°C=273.15 K and P=1 atm=101,3 kPa) one mole of real gas (especially in noble gases such as Argon) will behave like an ideal gas and the constant R will be [tex]8.3144598 \frac{Pa. m^{3}}{mol. K}[/tex].
However, in this case we are not working with standard temperature and pressure, therefore, even if we are working with Argon, the value of R will be far from the constant of the ideal gases.
Having this clarified, let's isolate [tex]R[/tex] from (1):
[tex]R=\frac{PV}{nT}[/tex] (2)
Where:
[tex]P=259392 Pa[/tex]
[tex]n=1 mole[/tex]
[tex]T=200\°C=473.15 K[/tex] is the absolute temperature of the gas in Kelvin.
[tex]V=8.8 dm^{3}=0.0088 m^{3}[/tex]
[tex]R=\frac{(259392 Pa)(0.0088 m^{3})}{(1 mole)(473.15 K)}[/tex] (3)
Finally:
[tex]R=4.82436 \frac{Pa. m^{3}}{mol. K}[/tex]