The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P.A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters. Find the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K. Round your answer to two decimal places. The pressure is Number atmospheres. Show your work and explain, in your own words, how you arrived at your answer. Answers with no relevant explanations may receive reduced or no credit. K ΞΕΩΣ GGGQE: 三 A- A-T BIUS X, x² Styles Size Font

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### Ideal Gas Law Problem for Educational Website

**Problem Statement:**

The volume \( V \) of an ideal gas varies directly with the temperature \( T \) and inversely with the pressure \( P \). A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters. Find the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K.

Round your answer to two decimal places.

**The pressure is** __________ atmospheres.

**Instructions:**
Show your work and explain, in your own words, how you arrived at your answer. Answers with no relevant explanations may receive reduced or no credit.

**Input Box:**
[Insert text box with the label "The pressure is (Number) atmospheres."]

**Solution Guidelines:**

1. **Understanding the Relationship:**
    - According to the problem, the volume \( V \) of an ideal gas is directly proportional to its temperature \( T \) and inversely proportional to its pressure \( P \). Mathematically, we can express this as:
      \[
      V = k \frac{T}{P}
      \]
      where \( k \) is a constant of proportionality.

2. **Determining the Constant:**
    - Substitute the initial conditions to find the constant \( k \):
      \[
      120 = k \frac{310}{18}
      \]
    - Solving for \( k \):
      \[
      k = 120 \times \frac{18}{310} = 6.9677
      \]

3. **Using the Constant:**
    - With \( k \) known, the relationship can be used to determine the new pressure when the volume is reduced to 100 liters and the temperature is increased to 320 degrees K:
      \[
      100 = 6.9677 \frac{320}{P}
      \]

4. **Solving for the New Pressure:**
    - Rearranging the equation to solve for \( P \):
      \[
      100P = 6.9677 \times 320
      \]
      \[
      P = \frac{6.9677 \times 320}{100} = 22.27 \text{ atmospheres}
      \]

Therefore, the correct answer is 22.27 atmospheres.

**End
Transcribed Image Text:### Ideal Gas Law Problem for Educational Website **Problem Statement:** The volume \( V \) of an ideal gas varies directly with the temperature \( T \) and inversely with the pressure \( P \). A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters. Find the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K. Round your answer to two decimal places. **The pressure is** __________ atmospheres. **Instructions:** Show your work and explain, in your own words, how you arrived at your answer. Answers with no relevant explanations may receive reduced or no credit. **Input Box:** [Insert text box with the label "The pressure is (Number) atmospheres."] **Solution Guidelines:** 1. **Understanding the Relationship:** - According to the problem, the volume \( V \) of an ideal gas is directly proportional to its temperature \( T \) and inversely proportional to its pressure \( P \). Mathematically, we can express this as: \[ V = k \frac{T}{P} \] where \( k \) is a constant of proportionality. 2. **Determining the Constant:** - Substitute the initial conditions to find the constant \( k \): \[ 120 = k \frac{310}{18} \] - Solving for \( k \): \[ k = 120 \times \frac{18}{310} = 6.9677 \] 3. **Using the Constant:** - With \( k \) known, the relationship can be used to determine the new pressure when the volume is reduced to 100 liters and the temperature is increased to 320 degrees K: \[ 100 = 6.9677 \frac{320}{P} \] 4. **Solving for the New Pressure:** - Rearranging the equation to solve for \( P \): \[ 100P = 6.9677 \times 320 \] \[ P = \frac{6.9677 \times 320}{100} = 22.27 \text{ atmospheres} \] Therefore, the correct answer is 22.27 atmospheres. **End
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