Chemistry, 20.01.2021 01:50 KylaChanel4756

A sample of gas has a volume of 2.00 L and a pressure of 0.750 kPa when its temperature is 25°C. If the volume is expanded to 4.00 L and the pressure reduced to

0.500 kPa, what must the temperature become?

379°C

397°C

379 K

397K

Answers: 2

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Answer from: eloyjrcruz101

The final temperature is 397 K.

Explanation:

Boyle's law relates volume and pressure by expressing:

"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

where k is a constant.

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 states: "At constant pressure, the volume that a gas sample occupies is directly proportional to the temperature"

So, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between the volume and the temperature will always have the same value:

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

Finally, Gay-Lussac's law establishes the relationship between pressure and temperature of a gas when the volume is constant. This law states that the pressure of a fixed volume of a gas is directly proportional to its temperature. That is, if the volume of a certain quantity of ideal gas at moderate pressure is kept constant, the quotient between pressure and temperature remains constant.

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

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

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

Supposing you have a gas that is at a pressure P1, at a volume V1 and at a temperature T1 at the beginning of the experiment, by varying the temperature to a new value T2, then the pressure will change to P2 and the volume to V2, and will fulfill:

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

In this case, you have:

P1= 0.750 kPaV1= 2 LT1= 25°C= 298 °K (being 0°C=273°K)P2= 0.500 kPaV2= 4 LT2= ?

Replacing:

$\frac{0.750 kPa* 2 L}{298 K}=\frac{0.500 kPa*4 L}{T2}$

and solving you get:

$T2=\frac{0.500 kPa*4 L}{\frac{0.750 kPa* 2 L}{298 K}}$

T2= 397 K

The final temperature is 397 K.

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