Question

A gas occupies a volume of 500. Milliliters at a pressure of 38.0 kPa and a temperature of 298 K. At what temperature will the gas occupy a volume of 250. Milliliters and have a pressure of 76.0 kPa?

Answer 1

Answer:

A gas occupies a volume of 250 mL at a pressure of 76 kPa and a temperature of 298 K.

Explanation:

Charles's law states that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. This mathematical law expresses that the quotient that exists between the volume and the temperature will always have the same value:

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

Gay-Lussac's law says 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 mathematically says that the quotient between pressure and temperature is constant:

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

Finally, Boyle's law says that the volume occupied by a certain gaseous mass at constant temperature is inversely proportional to the pressure. This law is expressed mathematically as

P * V = k

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

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

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

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

In this case:

• P1= 38 kPa
• V1= 500 mL
• T1= 298 K
• P2= 76 kPa
• V2= 250 mL
• T2= ?

Replacing:

$\frac{38 kPa*500 mL}{298 K} =\frac{76 kPa*250 mL}{T2}$

Solving:

$T2*\frac{38 kPa*500 mL}{298 K} =76 kPa*250 mL$

$T2=\frac{76 kPa*250 mL*298 K}{38 kPa*500 mL}$

T2= 298 K

A gas occupies a volume of 250 mL at a pressure of 76 kPa and a temperature of 298 K.