Given :
A constant volume of gas at 1.00 atm is heated from 30.0°C to 40.0°C.
To Find :
The pressure change.
Solution :
By Gay-Lussac's Law :
[tex]\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\\\dfrac{1}{30+273}=\dfrac{P_2}{40+273}\\\\P=\dfrac{313}{303}\\\\P=1.033\ atm[/tex]
Change in pressure,
[tex]\Delta \ P=P_2-P_1\\\\\Delta \ P=(1.033-1)\ atm\\\\\Delta \ P=0.33\ atm[/tex]
Therefore, the pressure change is 0.33 atm.
Hence, this is the required solution.
The change in pressure when a constant volume of gas at 1.00 atm is heated from 30.0°C to 40.0°C is 0.033 atm and this can be determined by using the Gay-Lusac's law.
Given :
Gay-Lussac's law can be used in order to determine the pressure change when a constant volume of gas at 1.00 atm is heated from 30.0°C to 40.0°C.
According to Gay-Lusac's law:
[tex]\rm \dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}[/tex]
Now, substitute the values of the known terms in the above expression.
[tex]\rm \dfrac{1}{30+273}=\dfrac{P_2}{40+273}[/tex]
Simplify the above expression in order to determine the value of [tex]\rm P_2[/tex].
[tex]\rm P_2 = \dfrac{313}{303}=1.033\;atm[/tex]
Now, the change in pressure is given by:
[tex]\rm \Delta P = P_2-P_1[/tex]
[tex]\rm \Delta P = 1.033-1=0.033\;atm[/tex]
Therefore, the change in pressure when a constant volume of gas at 1.00 atm is heated from 30.0°C to 40.0°C is 0.033 atm.
For more information, refer to the link given below:
https://brainly.com/question/6590381
