A gas is placed in a 5.0-mL syringe. It exerts 115 mm Hg of pressure on the inside walls of the syringe. The syringe's plunger is pressed, reducing the volume of the syringe to 2.1 mL. The cap was not removed from the syringe, so none of the gas escapes. Assuming the temperature of the gas does not change, use Boyle's law (below) to determine the pressure of the compressed gas.

273.8mmHg

Explanation:

Data obtained from the question include:

V1 (initial volume) = 5mL

P1 (initial pressure) = 115 mmHg

V2 (final volume) = 2.1 mL

P2 (final pressure) =?

Using the Boyle's law equation P1V1 = P2V2, the final pressure in the syringe can be obtained as follow:

P1V1 = P2V2

115 x 5 = P2 x 2.1

Divide both side by 2.1

P2 = (115 x 5) /2.1

P2 = 273.8mmHg

The final pressure i.e the pressure of the compressed gas is 273.8mmHg

The new pressure of the gas is 0.360 atm or 273.8 mmHg

Explanation:

Step 1: Data given

Volume = 5.0 mL = 0.005 L

The pressure = 115 mmHg = 115 / 760 atm = 0.151316 atm

The volume reduced to 2.1 mL  = 0.0021 L

Step 2: Calculate the new pressure of the gas

P1*V1 = P2*V2

⇒ with P1 = the initial pressure of the gas = 115 mmHg =  0.151316 atm

⇒ with V1 = the initial volume of the syringe = 5.0 mL = 0.005 L

⇒ with P2 = the final pressure of the gas = TO BE DETERMINED

⇒with V2 = the reduced volume = 2.1 mL = 0.0021 L

0.151316 * 0.005 = P2 * 0.0021

P2 = (0.151316 * 0.005) / 0.0021

P2 =  0.360 atm

OR

115 * 0.005 = P2 * 0.0021

P2 = 273.8 mmHg

The new pressure of the gas is 0.360 atm or 273.8 mmHg

more info ? ? i don't understand what the question is

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