Question

Sample of gas initially occupies 4.25 L at a pressure of 0.850 atm at 23.0°C. What will the volume be if the temperature is changed to 11.5°C, and the pressure is changed to 1.50 atm?

Answer 1

Answer:  2.34 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 = 0.850 atm

$P_2$ = final pressure of gas = 1.50 atm

$V_1$ = initial volume of gas = 4.25 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $23.0^oC=273+23.0=293.0K$

$T_2$ = final temperature of gas = $11.5^oC=273+11.5=284.5K$

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

$\frac{0.850\times 4.25}{293.0K}=\frac{1.50\times V_2}{284.5}$

$V_2=2.34L$

Thus the final volume will be 2.34 L