The volume of a gas is \(5.80 \mathrm{~L}\), measured at 1.00 atm. What is the pressure of the gas in \(\mathrm{mmHg}\) if the volume is changed to \(9.65 \mathrm{~L} ?\) (The temperature remains constant.)

Short Answer

Expert verified
The pressure of the gas, when the volume is changed to \(9.65 L\), is approximately \(455.17 mmHg\).

Step by step solution

01

Identify known and unknown quantities

Given values are: \n- Initial volume, \(V1 = 5.80 L\) \n- Initial pressure, \(P1 = 1.00 atm\) \n- Final Volume, \(V2 = 9.65 L\) \n\nWe need to find the final pressure \(P2\) in 'mmHg'). It's also important to note there are conversion factors that link 'atm' and 'mmHg': \(1 atm = 760 mmHg\).
02

Apply Boyle's law

Apply Boyle's Law, which is \(P1 \cdot V1 = P2 \cdot V2\) and rearrange to find \(P2\): \n\[P2 = \frac{P1 \cdot V1}{ V2}\] \n\nSubstitute given values into the formula: \n\n\[P2 = \frac{1.00 atm \cdot 5.80 L}{ 9.65 L}\]
03

Convert units and solve

Solve the equation to get the final pressure in atm, then convert the pressure from atm to mmHg using the conversion factor (\(1 atm = 760 mmHg\)):\n\n\[P2 = \frac{1.00 atm \cdot 5.80 L}{9.65 L}\] in atm, then \n\n\[P2 = P2_{atm} \cdot 760 \frac{mmHg}{atm}\] in mmHg.

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Most popular questions from this chapter

A certain amount of gas at \(25^{\circ} \mathrm{C}\) and at a pressure of 0.800 atm is contained in a glass vessel. Suppose that the vessel can withstand a pressure of 2.00 atm. How high can you raise the temperature of the gas without bursting the vessel?

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