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.

Answers

Answer 1

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 LInitial pressure (P₁) = 1 atmTemperature = 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 LInitial pressure (T₁) = 300 KPressure = 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 atmWhen 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.

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Answer 2

Answer:

b

Explanation:


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good luck

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Answers

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Further explanation

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Required

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i've taken the test

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Answers

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The final pressure in the flasks (containing N₂ and CO) after the stopcock is opened is 1.57 atm

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Solid Snake (75 kg) carries 18 kg of equipment. If Solid Snake climbs 27 flights of stairs, each 5.5 m in height, the work done against gravity is 135.34 kJ (c).

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In physics, work is the energy transferred to or from an object via the application of force along a displacement.

In this problem, the force that does the work is the one that opposes the weight.

Step 1: Calculate the total weight of the system.

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We can calculate the weight using Newton's second law of motion.

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