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The volume of a gas is 3.0 L, the pressure is 1.0 atm, and the temperature is 300 K. A chemist changes one factor while keeping another constant so that the new volume is 2.0 L. Which of the following could be the new conditions?

Group of answer choices

The final pressure is 0.5 atm, while temperature is kept constant.

The final temperature is 200 K, while pressure remains constant.

The final pressure is 0.2 atm, while temperature is kept constant.

The final temperature is 350 K, while pressure remains constant.

Respuesta :

Eduard22sly

Given the data from the question, the final temperature is 200 K, while pressure remains constant.

Basic concepts

To obtain the correct answer to the question, we shall consider two conditions:

  • Case 1 (temperature is constant)
  • Case 2 (pressure is constant)

Case 1 (Temperature is constant)

We shall determine the new pressure by using the combined gas equation (P₁V₁ / T₁ = P₂V₂ / T₂) as illustrated below:

  • Initial volume (V₁) = 3 L
  • Initial pressure (P₁) = 1 atm
  • Temperature = constant
  • New Volume (V₂) = 2 L
  • New pressure (P₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

Since temperature is constant, we have:

P₁V₁ = P₂V₂

3 × 1 = P₂ × 2

3 = P₂ × 2

Divide both side by 2

P₂ = 3 / 2

P₂ = 1.5 atm

Case 2 ( pressure is constant)

We shall determine the new temperature by using the combined gas equation (P₁V₁ / T₁ = P₂V₂ / T₂) as illustrated below:

  • Initial volume (V₁) = 3 L
  • Initial pressure (T₁) = 300 K
  • Pressure = constant
  • New Volume (V₂) = 2 L
  • New pressure (T₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

Since pressure is constant, we have:

V₁ / T₁ = V₂ / T₂

3 / 300 = 2 / T₂

1 / 100 = 2 / T₂

Cross multiply

T₂ = 100 × 2

T₂ = 200 K

SUMMARY

  • when the temperature is constant, the new pressure is 1.5 atm
  • When the pressure is constant, the new temperature is 200 K

From the calculations made above, we can conclude that the correct answer is:

The final temperature is 200 K, while pressure remains constant.

Learn more about gas laws:

https://brainly.com/question/6844441

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