Question

18. A balloon contains 7.2 L of He. The pressure is reduced to
|| 2.00 atm and the
balloon expands to occupy a volume of 25.1 L. What was
the initial pressure
exerted on the balloon?​

Answer 1

Answer:

The initial pressure exerted on the balloon is 7.0 atm.

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

Initial pressure of helium gas in balloon = $P_1=?$

Initial volume of helium gas in balloon = $V_1=7.2 L$

Final pressure of helium gas in balloon = $P_2=2.00 atm$

Final volume of helium gas in balloon = $V_2=25.1 L$

Putting values in above equation, we get:

$P_1\times 7.2 L=2.00 atm\times 25.1 L$

$P_1=\frac{2.00 atm\times 25.1 L}{7.2 L}=6.97 atm\approx 7.0 atm$

Hence, the initial pressure exerted on the balloon is 7.0 atm.