190k views
0 votes
The air inside a tire pump occupies a volume of 130. cm^3 at a pressure of one atmosphere. If the volume decreases to 40.0 cm^3, what is the pressure, in atmospheres, inside the pump?

User Gbps
by
6.2k points

1 Answer

5 votes

the Given data:

Initial volume;


V_1=130\operatorname{cm}^3

Initial pressure;


P_1=1\text{ atm}

Final volume;


V_2=40\operatorname{cm}^3

According to ideal gas equation;


PV=\text{constant}

Therefore,


P_1V_1=P_2V_2^{}

Here, P_2 is the final pressure.

The final pressure is given as,


P_2=(P_1V_1)/(V_2)

Substituting all known values,


\begin{gathered} P_2=\frac{(1\text{ atm})*(130\operatorname{cm})}{(40\operatorname{cm}^3)} \\ =3.25\text{ atm} \end{gathered}

Therefore, the new pressure inside the pump is 3.25 atmospheres.

User CyberGuy
by
7.0k points