Question

A gas balloon has a volume of 80.0 mL at 300K, and a pressure of 50.0 kPa. If the pressure changes to 80 kPa and the temperature rises to 320K, what is the new volume of the balloon in millimeters?

Answer 1

Answer:  The new volume is 53.3 ml

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 =  50.0 kPa

$P_2$ = final pressure of gas = 80.0 kPa

$V_1$ = initial volume of gas = 80.0 ml

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $300K$

$T_2$ = final temperature of gas = $320K$

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

$\frac{50.0\times 80.0}{300}=\frac{80.0\times V_2}{320}$

$V_2=53.3ml$

The new volume is 53.3 ml