A certain gas is present in a 11.0 \({\rm L}\) cylinder at 1.0 \({\rm atm}\) pressure. If the pressure is increased to 2.0 \({\rm atm}\) , the volume of the gas decreases to 5.5 \({\rm L}\) . Find the two constants \(k_{\rm i}\), the initial value of \(k\), and \(k_{\rm f \hspace{1 pt}}\), the final value of \(k\), to verify whether the gas obeys Boyle’s law.

Express your answers to two significant figures separated by a comma

does 11.0 * 1.0 = 5.5 * 2.0 ?

To verify whether the gas obeys Boyle's law, we need to calculate the initial and final values of the constant \(k\) and compare them.

Boyle's law states that the product of pressure and volume of a gas sample remains constant at a constant temperature. Mathematically, it can be expressed as:

\(P_1 \cdot V_1 = P_2 \cdot V_2\)

Where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume.

In this case, we are given:

Initial volume \(V_1 = 11.0 \, \rm L\)
Initial pressure \(P_1 = 1.0 \, \rm atm\)

Final volume \(V_2 = 5.5 \, \rm L\)
Final pressure \(P_2 = 2.0 \, \rm atm\)

We can rearrange the Boyle's law equation to solve for \(k\):
\(k = \frac{{P \cdot V}}{T}\)

For the initial state, \(k_{\rm i} = P_1 \cdot V_1\) and for the final state, \(k_{\rm f} = P_2 \cdot V_2\).

Let's calculate these values:

\(k_{\rm i} = 1.0 \, \rm atm \times 11.0 \, \rm L\)
\(k_{\rm f} = 2.0 \, \rm atm \times 5.5 \, \rm L\)

Calculating these values gives:

\(k_{\rm i} = 11.0 \, \rm atm \cdot L\)
\(k_{\rm f} = 11.0 \, \rm atm \cdot L\)

As we can see, both \(k_{\rm i}\) and \(k_{\rm f}\) are equal to 11.0 atm·L. This indicates that the gas obeys Boyle's law because the value of \(k\) remains constant before and after the change in pressure and volume.

Therefore, the constants \(k_{\rm i}\) and \(k_{\rm f}\) are both 11.0 atm·L.