Question

1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressure of the gas?
2. If a gas at 25.0 °C occupies 3.60 liters at a pressure of 10 kPa, what will be its volume at a pressure of 25 kPa?



3. When the pressure on a gas increases three times, by how much will the volume increase or decrease?


4. Boyle's Law deals what quantities?

Answer 1

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$, $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$, then the new pressure of the gas is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 500\,kPa$

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $V_{1} = 3.60\,L$, $P_{1} = 10\,kPa$ and $P_{2} = 25\,kPa$ then the new volume of the gas is:

$V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)$

$V_{2} = 1.44\,L$

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $\frac{P_{2}}{P_{1}} = 3$, then the volume ratio is:

$\frac{V_{1}}{V_{2}} = 3$

$\frac{V_{2}}{V_{1}} = \frac{1}{3}$

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.