Question

(II) If 61.5 L of oxygen at 18.0°C and absolute pressure of 2.45 atmis compressed to 38.8 L, and at the same time, the temperature is raised to 56.0°C, what will the new pressure be?

Short answer

The final temperature of the oxygen gas is \(4.39\;{\rm{atm}}\).

Step-by-step solution

Concepts

The ideal gas law is \(PV = nRT\) for n mole of ideal gas.

For this problem, you have to use the equation\(\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}\).

Given data 

The initial volume of the oxygen is\({V_1} = 61.5\;{\rm{L}}\).

The final volume of the oxygen is \({V_2} = 38.8\;{\rm{L}}\).

The initial pressure is \({P_1} = 2.45\;{\rm{atm}}\).

The initial temperature is \({T_1} = {18.0^ \circ }{\rm{C}} = 291\;{\rm{K}}\).

The final temperature is \({T_2} = {56.0^ \circ }{\rm{C}} = 329\;{\rm{K}}\).

Let \({P_2}\) be the final pressure of the oxygen gas.

Calculation 

Now from the ideal gas law, you getthe following:

\(\begin{array}{c}\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}\\{P_2} = {P_1} \times \frac{{{T_2}}}{{{T_1}}} \times \frac{{{V_1}}}{{{V_2}}}\end{array}\)

After further calculation,you get the following:

\(\begin{array}{c}{P_2} = \left( {2.45\;{\rm{atm}}} \right) \times \frac{{329\;{\rm{K}}}}{{291\;{\rm{K}}}} \times \frac{{61.5\;{\rm{L}}}}{{38.8\;{\rm{L}}}}\\{P_2} = 4.39\;{\rm{atm}}\end{array}\)

Hence, the final temperature of the oxygen gas is \(4.39\;{\rm{atm}}\).