Hello!
Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 C to 30.0 C
We have the following information:
P1 (initial pressure) = 1 atm
T1 (initial temperature) = 20 ºC (in Kelvin)
TK = TºC+273.15 → TK = 20+273.15 → T1 (initial temperature) = 293.15 K
P2 (final pressure) = ? (in atm)
T2 (final temperature) = 30 ºC (in Kelvin)
TK = TºC+273.15 → TK = 30+273.15 → T2 (final temperature) = 303.15 K
According to the Law of Charles and Gay-Lussac in the study of gases, we have an isochoric (or isovolumetric) transformation when its volume remains constant or equal, if the temperature increases, the pressure increases and, if the temperature decreases, the pressure also decreases, they are directly proportional.then we will have the following formula:
[tex]\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}[/tex]
[tex]\dfrac{1}{293.15} = \dfrac{P_2}{303.15}[/tex]
multiply the means by the extremes
[tex]293.15*P_2 = 1*303.15[/tex]
[tex]293.15\:P_2 = 303.15[/tex]
[tex]P_2 = \dfrac{303.15}{293.15}[/tex]
[tex]\boxed{\boxed{P_2 \approx 1.03\:atm}}\end{array}}\qquad\checkmark[/tex]
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I Hope this helps, greetings ... Dexteright02! =)
Pressure is the force experienced by the surface and is determined in pascals or atm. The pressure change when the system is heated is 1.03 atm.
They gave the law when the gas of the system is in the isochoric condition that is the volume is constant. The pressure and the temperature follows the direct proportionality rule and is given as,
[tex]\rm \dfrac{P_{1}}{T_{1}} = \rm \dfrac{P_{2}}{T_{2}}[/tex]
Where,
Initial pressure = 1 atm
Initial temperature = 293.15 K
Final temperature = 303.15 K
Substituting values in the above equation:
[tex]\begin{aligned}\dfrac{1}{293.15} &= \dfrac{\rm P_{2}}{303.15}\\\\\rm P_{2} & = \dfrac{303.15}{293.15}\\\\&= 1.03\;\rm atm\end{aligned}[/tex]
Therefore, the pressure change is 1.03 atm.
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