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Angelina_Jolie [31]
3 years ago
13

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?

Chemistry
1 answer:
Sonja [21]3 years ago
7 0

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,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.892 atm

P_2 = final pressure of gas = 1.05 atm

V_1 = initial volume of gas = 6.35 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 55.0^oC=273+55.0=328K

T_2 = final temperature of gas = 59.0^oC=273+59.0=332K

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

\frac{0.892atm\times 6.35L}{328K}=\frac{1.05atm\times V_2}{332K}

V_2=5.46L

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

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