Question

A sample of oxygen occupies 560. when the pressure is 106.7 kPa. At a constant temperature, what volume in does the gas occupy when the pressure decreases to 659 mm Hg?

Answer 1

To solve this problem, we can use the combined gas law equation, which relates the initial and final volumes, pressures, and temperatures of a gas sample:

P1V1 / T1 = P2V2 / T2

Where:

P1 = initial pressure = 106.7 kPa

V1 = initial volume = 560 mL

T1 = constant temperature (not given)

P2 = final pressure = 659 mmHg

V2 = final volume (unknown)

Before we can use this equation, we need to convert the units of pressure to the same system. Let's convert the initial pressure from kPa to mmHg:

106.7 kPa * 760 mmHg / 101.3 kPa = 800 mmHg

Now we can plug in the values and solve for V2:

800 mmHg * 560 mL / T1 = 659 mmHg * V2 / T1

We can simplify this equation by canceling out the T1 terms on both sides, and then solving for V2:

V2 = (800 mmHg * 560 mL) / 659 mmHg

V2 = 679 mL (rounded to three significant figures)

Therefore, the gas sample occupies 679 mL at a pressure of 659 mmHg, assuming constant temperature.

Step-by-step explanation:

Answer 2

Answer: The volume of the gas when the pressure decreases to 659 mm Hg (or 87.84 kPa) is approximately 678.9 liters.

Step-by-step explanation:

To solve this problem, we can use the principle of Boyle's Law, which states that the pressure of a given amount of gas held at constant temperature is inversely proportional to the volume of the gas. In other words, if temperature is constant, P1V1 = P2V2, where:

- P1 and V1 are the initial pressure and volume

- P2 and V2 are the final pressure and volume

Given in the problem:

- P1 = 106.7 kPa

- V1 = 560.0 L

- P2 = 659 mm Hg

First, we need to make sure that our pressures are in the same units. We can convert mm Hg to kPa (since 1 kPa = 7.50062 mm Hg):

P2 = 659 mm Hg / 7.50062 mm Hg/kPa = 87.84 kPa

Then we can substitute these values into Boyle's Law to solve for V2:

P1V1 = P2V2

106.7 kPa * 560.0 L = 87.84 kPa * V2

V2 = (106.7 kPa * 560.0 L) / 87.84 kPa

Let's compute the value of V2.

After performing the calculation, we find:

V2 = (106.7 kPa * 560.0 L) / 87.84 kPa = 678.9 L

So, the volume of the gas when the pressure decreases to 659 mm Hg (or 87.84 kPa) is approximately 678.9 liters.