Initial volume is 2.3L, initial temperature 150°C and initial pressure 2.5atm. Temperature increases to 300°C and pressure doubles. The new volume is?

Respuesta :

This is using the Combined Gas Law, which is the following:

[tex] \frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}[/tex], or [tex]P_{1} V_{1} T_{2} = P_{2} V_{2} T_{1} [/tex]

Since you are solving for [tex] V_{2} [/tex], you can change the equation to be: [tex] V_{2} [/tex] = [tex] \frac{P_{1} V_{1} T_{2}}{{P_2}{T_1}} [/tex]

Your given values are as follows:
[tex] P_{1} [/tex]: 2.5atm
[tex] V_{1} [/tex]: 2.3L
[tex] T_{1} [/tex]: 150°C
[tex] P_{2} [/tex]: 5atm
[tex] T_{2} [/tex]: 300°C

This formula only works if temperature is given in terms of Kelvin, so you can convert 150°C and 300°C to Kelvin by adding 273. Doing this, you get the following values:

[tex] T_{1} [/tex]=423K
[tex] T_{2} [/tex]573K

You then plug in all of the values you are given to get:

[tex] V_{2} [/tex] = [tex]\frac {(2.5atm)(2.3L)(423K)}{(5atm)(573K)}[/tex]

Multiply/divide all of the values to get:

[tex] V_{2} [/tex] = 0.85L rounded to the nearest hundredth