Question

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?

Answer 1

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