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-Dominant- [34]
3 years ago
7

A sample of gas has a volume of 2.00 L and a pressure of 0.750 kPa when its

Chemistry
1 answer:
Gennadij [26K]3 years ago
4 0

Answer:

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 kPa
  • V1= 2 L
  • T1= 25°C= 298 °K (being 0°C=273°K)
  • P2= 0.500 kPa
  • V2= 4 L
  • T2= ?

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

<u><em>The final temperature is 397 K.</em></u>

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