Respuesta :
The final pressure of the gas at 248 K is 0.908 atm
Explanation:
At standard temperature and pressure, temperature of 0 Celsius and pressure which is equal to the atmosphere, 1 atm.
[tex]{T_{1}}[/tex] = 0 Celsius
[tex]{P_{1}}[/tex] = 1 atm
Change in temperature causes a change in the pressure:
[tex]{T_{2}}[/tex]=248K
[tex]{P_{2}}[/tex]=?
Calculating the pressure, [tex]{P_{2}}[/tex] using the Gay-Lusaac's law:
[tex]\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}[/tex]
Convert the temperatures from Celsius to Kelvin:
[tex]{T_{1}}[/tex]=0 Celsius+273=273K
[tex]{T_{2}}[/tex]=248K
To find [tex]{P_{2}}[/tex]:
[tex]\frac{1 \text { atm }}{273 k}=\frac{P_{2}}{248 k}[/tex]
[tex]{P_{2}}[/tex]=[tex]\frac{248}{273}[/tex] = 0.908 atm
At the standard temperature and pressure, the experimental values can be compared. The final pressure of the gas at STP will be 0.908 atm.
What is Gay-Lusaac's law?
Gay-Lusaac's stated the relationship between the ratio of the initial pressure and temperature with that of the final. Gay-Lusaac's formula can be given as,
[tex]\rm \dfrac{P_{1}}{T_{1}} = \rm \dfrac{P_{2}}{T_{2}}[/tex]
At, STP pressure is 1 atm, the temperature is 273 K and the final temperature is 248 K. The pressure is calculated as:
[tex]\begin{aligned} \rm P_{2} &= \rm \dfrac{P_{1}}{T_{1}} \times T_{2}\\\\&= \dfrac{248}{273}\\\\&= 0.908 \;\rm atm\end{aligned}[/tex]
Therefore, at STP the final pressure is 0.908 atm.
Learn more about STP here:
https://brainly.com/question/15524102
