A 6.35 l sample of carbon monoxide is collected at 55.0◦c and 0.892 atm. What volume will the gas occupy at 1.05 atm and 59.0◦c?

Respuesta :

Answer : The final volume of gas will be, 5.46 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 = 0.892 atm

[tex]P_2[/tex] = final pressure of gas = 1.05 atm

[tex]V_1[/tex] = initial volume of gas = 6.35 L

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

[tex]T_1[/tex] = initial temperature of gas = [tex]55.0^oC=273+55.0=328K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]59.0^oC=273+59.0=332K[/tex]

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

[tex]\frac{0.892atm\times 6.35L}{328K}=\frac{1.05atm\times V_2}{332K}[/tex]

[tex]V_2=5.46L[/tex]

Thus, the final volume of gas will be, 5.46 L

The final volume of gas will be 5.46 L. This can be calculated by using combined gas law.

Combined gas law:

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

The combined gas equation is,

P₁*V₁/T₁ = P₂*V₂/T₂

where,

P₁ = initial pressure of gas = 0.892 atm

P₂= final pressure of gas = 1.05 atm

V₁ = initial volume of gas = 6.35 L

To find:

V₂ = final volume of gas = ?

V₁ = initial temperature of gas = 328 K

T = final temperature of gas = 332 K

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

(0.892 atm *6.35 L)/328 K= (1.05 atm * V₂)/ 332 K

V₂= 5.46 L

Find more information about Combined Gas law here:

brainly.com/question/25587265

Thus, the final volume of gas will be, 5.46 L