Question

A gas has three initial measurements: temperature 306 K, pressure 740 mm Hg, and volume 36 mL. The gas's temperature is increased to 850 K while its pressure is increased to 1750 mm Hg. What is the new volume of the gas in mL?

Answer 1

Answer:

The new volume of the gas is 42.28 mL.

Explanation:

Gay Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. This law can be expressed mathematically as follows:

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

where k = Constant

Boyle's law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This law is expressed mathematically as:

P * V = k

Finally, Charles's Law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

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

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

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

Studying an initial state 1 and a final state 2, it is satisfied:

$\frac{P1*V1}{T1}=\frac{P2*V2}{T2}$

In this case:

• P1= 740 mmHg
• V1= 36 mL
• T1= 306 K
• P2= 1750 mmHg
• V2= ?
• T2= 850 K

Replacing:

$\frac{740 mmHg*36 mL}{306 K}=\frac{1750 mmHg*V2}{850 K}$

Solving:

$V2=\frac{850 K}{1750 mmHg} *\frac{740 mmHg*36 mL}{306 K}$

V2= 42.28 mL

The new volume of the gas is 42.28 mL.