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Chapter 5: Problem 21

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 changes to \(9.65 \mathrm{~L}\) is approximately \(456.71 \mathrm{~mmHg}\).

Step by step solution

01

Identify Given Values

The initial volume (\(V鈧乗)) of the gas is given as \(5.80 \mathrm{~L}\), the initial pressure (\(P鈧乗)) as 1.00 atm and the final volume (\(V鈧俓)) as \(9.65 \mathrm{~L}\). The goal is to solve for the final pressure (\(P鈧俓)).
02

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 \(P鈧乗) in mmHg is \(1.00 \times 760 = 760 \mathrm{~mmHg}\).
03

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 \(P鈧乂鈧� = P鈧俈鈧俓). It is required to find \(P鈧俓) which can be done by rearrangement of the equation as \(P鈧� = \frac{P鈧乂鈧亇{V鈧倉\).
04

Substitute Values and Calculate

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

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