Question

10. A 10.9ml sample of gas is collected over water at 21.0 C and 1.89 atm , what volume would you have at 25.0 C and 2.25 atm

Answer 1

Considering the combined law equation, you would have a volume of 9.28 mL at 25.0 C and 2.25 atm.

Boyle's law

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant.

This law says that volume is inversely proportional to pressure: if pressure increases, volume decreases, while if pressure decreases, volume increases.

Mathematically, Boyle's law states that the multiplication of pressure by volume is constant:

P×V= k

Charles's law

Charles's law establishes the relationship between the temperature and the volume of a gas when the pressure is constant.

This law says that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles' law states that the ratio of volume to temperature is constant:

$\frac{V}{T}=k$

Gay-Lussac's law

Gay-Lussac's law establishes the relationship between the temperature and pressure of a gas when the volume is constant.

This law says that the pressure of the gas is directly proportional to its temperature: if the temperature increases, the pressure increases, while if the temperature of the gas decreases, the pressure decreases.

Mathematically, Gay-Lussac's law states that the ratio of pressure to temperature is constant:

$\frac{P}{T}=k$

Combined law equation

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

$\frac{PV}{T}=k$

Considering an initial state 1 and a final state 2, the following holds:

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

Volume in this case

In this case, you know:

• P1= 1.89 atm
• V1= 10.9 mL
• T1= 21 C= 294 K (being 0 C= 273 K)
• P2= 2.25 atm
• V2= ?
• T2= 25 C= 298 K

Replacing in the combined law equation:

$\frac{1.89 atmx10.9 mL}{294 K}=\frac{2.25 atmxV2}{298 K}$

Solving:

$\frac{1.89 atmx10.9 mL}{294 K}\frac{298K}{2.25 atm} =V2$

9.28 mL= V2

Finally, you would have a volume of 9.28 mL at 25.0 C and 2.25 atm.

Learn more about combined law equation:

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