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A 6.87-L sample of gas has a pressure of 0.732 atm and a temperature of 95 °C. The sample is allowed to expand to a volume of 9.22 L and is cooled to 44 °C. Calculate the new pressure of the gas, assuming that no gas escaped during the experiment.

Respuesta :

Answer: The new pressure of the gas, assuming that no gas escaped during the experiment is 0.470 atm

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 = 0.732 atm

[tex]P_2[/tex] = final pressure of gas = ?

[tex]V_1[/tex] = initial volume of gas = 6.87 L

[tex]V_2[/tex] = final volume of gas = 9.22 L

[tex]T_1[/tex] = initial temperature of gas = [tex]95^oC=273+95=368K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]44^oC=273+44=317K[/tex]

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

[tex]\frac{0.732\times 6.87}{368}=\frac{P_2\times 9.22}{317}[/tex]

[tex]P_2=0.470atm[/tex]

Thus the new pressure of the gas, assuming that no gas escaped during the experiment is 0.470 atm

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