Question

Two gas-containers, A and B, are connected with a valve. The first container, A, has a volume of 135 mL, and the second container, B, has a volume of 117 mL. A sample of gas originally in container A is at 22.5 C and the pressure is 165 mmHg. What is the pressure (in mmHg) of the gas sample when the valve is opened and the gas now occupies both containers at a temperature of 12.7 C

Answer 1

Answer:

85.5 mmHg is the pressure of the gas sample when the valve is opened.

Explanation:

The combined gas equation is,

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

where,

$P_1$ = initial pressure of gas in container A = 165 mmHg

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas in container A= $135 mL$

$V_2$ = final volume of gas = 135 mL + 117 mL = 252 mL

$T_1$ = initial temperature of gas in container A = $22.5^oC=273+22.5=295.5 K$

$T_2$ = final temperature of gas = $12.7^oC=273+12.7=285.7K$

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

$P_2=\frac{P_1V_1\times T_2}{T_1\times V_2}$

$=\frac{165 mmHg\times 135 mL\times 285.7 K}{295.5 K\times 252 mL}$

$P_2=85.5 mmHg$

85.5 mmHg is the pressure of the gas sample when the valve is opened.