Question

What happens to the air pressure inside a balloon when it is squeezed to half its volume at constant temperature? What happens to the air pressure inside a balloon when it is squeezed to half its volume at constant temperature? It is cut in half. It doubles. It quadruples. It remains the same.

Answer 1

The correct option is option (B)

Therefore the air pressure inside the balloon will be double.

Step-by-step explanation:

Boyle's Law:

At a constant temperature, the pressure of a given mass of an ideal gas varies inversely to its volume.

`P\propto \frac1 V`

Charles Law:

At a constant pressure, the volume of a given mass of an ideal gas varies directly to its temperature (in kelvin).

`V\propto T`

Combination of two laws is

`(PV)/(T)=constant`

Then,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

Given that,

The final volume is half of initial volume and the temperature remains constant.

So,
`T_1=T_2`,
`V_2=\frac12 V_1`

Then,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

`\Rightarrow (P_1V_1)/(T_1)=(P_2.\frac12 V_1)/(T_1)`

`\Rightarrow P_1=\frac12 P_2`

`\Rightarrow P_2= 2P_1`

⇒Final Pressure = twice of the initial pressure

Therefore the air pressure inside the balloon will be double.