Question

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

The pressure of the gas when the volume changes to \(9.65 \mathrm{~L}\) is approximately \(456.71 \mathrm{~mmHg}\).

Step-by-step solution

Identify Given Values

The initial volume (\(V₁\)) of the gas is given as \(5.80 \mathrm{~L}\), the initial pressure (\(P₁\)) as 1.00 atm and the final volume (\(V₂\)) as \(9.65 \mathrm{~L}\). The goal is to solve for the final pressure (\(P₂\)).

Convert Pressure to Desired Unit

Since the final answer is required to be in mmHg, convert the initial pressure from atm to mmHg. 1 atm is approximately equal to 760 mmHg. Therefore, the initial pressure \(P₁\) in mmHg is \(1.00 \times 760 = 760 \mathrm{~mmHg}\).

Apply Boyle's Law

Applying the principle of Boyle's law, the relationship between the initial and final volumes and pressures can be expressed as \(P₁V₁ = P₂V₂\). It is required to find \(P₂\) which can be done by rearrangement of the equation as \(P₂ = \frac{P₁V₁}{V₂}\).

Substitute Values and Calculate

Substitute the given values into the equation from step 3 to solve for \(P₂\): \(P₂ = \frac{760 \mathrm{~mmHg} \times 5.80 \mathrm{~L}}{9.65 \mathrm{~L}}\). The value of \(P₂\) is approximately 456.71 \(\mathrm{mmHg}\).