Question

a gaseous air - fuel mixture in a sealed car engine cylinder has an initial volume of 600. ml at 1.0 atm. to prepare for ignition of the fuel, a piston moves within the cylinder, reducing the volume of the air - fuel mixture to 50. ml at constant temperature. assuming ideal behavior, what is the new pressure of the air - fuel mixture? mark for review a about 0.083 atm, because the initial volume of the gas mixture was 12 times larger than its final volume. b about 12 atm, because the volume of the gas mixture decreased by a factor of 12. c about 120. atm, because the initial volume of the gas mixture was 120 times larger than its final volume. d about 550. atm, because the change in the volume of the gas mixture was 550. ml.

Explanation:

1. Identify the relevant gas - law:

• We know that for a gas at constant temperature (\(T =\) constant), Boyle's law applies, which is \(P_1V_1 = P_2V_2\), where \(P_1\) and \(V_1\) are the initial pressure and volume, and \(P_2\) and \(V_2\) are the final pressure and volume.
• Given \(V_1=600\ mL\), \(P_1 = 1.0\ atm\), and \(V_2 = 50\ mL\).

1. Calculate the final pressure \(P_2\):

• From Boyle's law \(P_2=\frac{P_1V_1}{V_2}\).
• Substitute the values: \(P_2=\frac{1.0\ atm\times600\ mL}{50\ mL}=12\ atm\).
• The ratio of the final pressure to the initial pressure is \(\frac{P_2}{P_1}=\frac{12\ atm}{1\ atm}=12\). This means the pressure of the gas - fuel mixture is increased by a factor of 12.

Step1: Identify the gas - law

For a constant - temperature process, use Boyle's law \(P_1V_1 = P_2V_2\).

Step2: Calculate the final pressure

Substitute \(P_1 = 1.0\ atm\), \(V_1 = 600\ mL\), and \(V_2 = 50\ mL\) into \(P_2=\frac{P_1V_1}{V_2}\), so \(P_2=\frac{1.0\times600}{50}=12\ atm\).

Answer:

B. About 12, because the volume of the gas mixture decreased by a factor of 12.