Question

When extended, a bicycle pump has a volume of 0.952 L at STP. What is its pressure when the pump is compressed to a new volume of 0.225 L and 278 K? (Think Combined)

Answer 1

We can assume that the gas inside the bicycle pump does not change its number of moles. Furthermore we will assume the gas behaves as an ideal gas so it is described by the following equation:

`\begin{gathered} PV=nRT \\ n=(PV)/(RT) \end{gathered}`

Where,

P is the pressure of the gas

V is the volume of the gas

T is the temperature of the gas

n is the number of moles of the gas

R is a constant =0.08206 atm.L/ol.K

Now, we have two states, state 1 at STP conditions and state 2. Since the number of moles remains constant we can equate the two states.

`\begin{gathered} (P_1V_1)/(RT_1)=n=(P_2V_2)/(RT_2) \\ (P_1V_1)/(T_1)=(P_2V_2)/(T_2) \end{gathered}`

Now, the conditions for each state are:

State 1.

V1=0.952L

T1=273.15K (STP)

P1=1atm (STP)

State 2.

V2=0.225L

T2=278K

P2=?

We clear V2 and replace the known data:

`P_{2=\text{ }}(P_1V_1)/(T_1)*(T_2)/(V_2)`
`\begin{gathered} P_{2=\text{ }}(1atm*0.952L)/(273.15K)*(278K)/(0.225L) \\ P_{2=\text{ }}4.31atm \end{gathered}`

Answer: The new pressure will be 4.31atm