Question

A 850.0 mL cylinder containing B2H6 at a pressure of 0.900 atm is connected by a valve to 1000 mL cylinder containing C6H6 at 608.0 torr pressure. Calculate the partial pressure (torr) of C6H6 when the valve is opened.

Answer 1

Answer:

The partial pressure of $C_6H_6$ when the valve is opened is 328.6 Torr.

Explanation:

Initial pressure of the $C_6H_6=P_1=608.0 Torr$

Initial volume of the $C_6H_6=V_1=1000 mL$

After opening the valve :

Final pressure or partail pressure of the $C_6H_6=P_2$

Final volume of the $C_6H_6=V_2=1000 mL+850.0 mL=1850.0 mL$

$P_1V_1=P_2V_2$ ( Boyle's law)

$P_2=\frac{608.0 Torr \times 1000 mL}{1850.0 mL}=328.6 Torr$

The partial pressure of $C_6H_6$ when the valve is opened is 328.6 Torr.