Question

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

Answer 1

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

The final temperature is 397 K.

Answer 2

The new temperature of the gas is 397 K

Data obtained from the question

• Initial volume (V₁) = 2 L
• Initial pressure (P₁) = 0.750 KPa
• Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
• New Volume (V₂) = 4 L
• New pressure (P₂) = 0.500 KPa
• New temperature (T₂) =?

How to determine the new temperature

The new temperature of the gas can be obtained by using the combined gas equation as illustrated below:

P₁V₁ / T₁ = P₂V₂ / T₂

(0.75 × 2) / 298 = (0.5 × 4) / T₂

1.5 / 298 = 2 / T₂

Cross multiply

1.5 × T₂ = 298 × 2

1.5 × T₂ = 596

Divide both side by 1.5

T₂ = 596 / 1.5

T₂ = 397 K

Learn more about gas laws:

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