Question

At 50.7 kPa and 15.0 °C a sample of gas occupies 120 L. Using the Combined Gas Law, what pressure does it occupy at 236 L and 10.0°C?
38.7 kPa
25.3 kPa
17.2 kPa
26.2 kPa

Answer 1

Answer:

A volume of 236 L and a temperature of 10.0 ° C has a pressure of 25.3 kPa.

Explanation:

Boyle's law says that "The volume occupied by a certain gaseous mass at constant temperature is inversely proportional to pressure." Boyle's law is expressed mathematically as:  P * V = k

Gay-Lussac's law can be expressed mathematically as follows:

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

This 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.

Finally, Charles's Law consists of the relationship between the volume and the temperature of a certain amount of ideal gas, which is kept at a constant pressure. This law is a law that says that when the amount of gas and pressure are kept constant, the quotient 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 two different states, an initial state and a final state, you have:

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

In this case:

• P1= 50.7 kPa
• V1= 120 L
• T1= 15 C= 288 K (being 0 C= 273 K)
• P2= ?
• V2= 236 L
• T2= 10 C= 283 K

Replacing:

$\frac{50.7 kPa*120 L}{288 K} =\frac{P2*236 L}{283 K}$

Solving:

$P2=\frac{283 K}{236 L} *\frac{50.7 kPa*120 L}{288 K}$

P2= 25.3 kPa

A volume of 236 L and a temperature of 10.0 ° C has a pressure of 25.3 kPa.