Sample of gas initially occupies 4.25 L at a pressure of 0.850 atm at 23.0°C. What will the volume be if the temperature is changed to 11.5°C, and the pressure is changed to 1.50 atm?

Respuesta :

Answer:  2.34 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 0.850 atm

[tex]P_2[/tex] = final pressure of gas = 1.50 atm

[tex]V_1[/tex] = initial volume of gas = 4.25 L

[tex]V_2[/tex] = final volume of gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]23.0^oC=273+23.0=293.0K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]11.5^oC=273+11.5=284.5K[/tex]

Now put all the given values in the above equation, we get:

[tex]\frac{0.850\times 4.25}{293.0K}=\frac{1.50\times V_2}{284.5}[/tex]

[tex]V_2=2.34L[/tex]

Thus the final volume will be 2.34 L

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