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Chapter 11: Problem 116

A gas tank contains \(\mathrm{CO}_{2}\) at a pressure of \(6.80 \mathrm{~atm}\). What would the \(\mathrm{CO}_{2}\) pressure be if the container were (a) twice as large and (b) one-fourth as large?

Short Answer

Expert verified
The pressure of CO₂ when the container is (a) twice as large is \(3.40 \mathrm{~atm}\) and (b) one-fourth as large is \(27.20 \mathrm{~atm}\).

Step by step solution

01

Identify the variables and write the Boyle's Law equation

Let the initial volume of the gas tank be V₁, and the initial pressure be P₁. Let the new volumes and pressures be V₂ and P₂, respectively. According to Boyle's Law, the product of pressure and volume remains constant for a given amount of gas at a constant temperature. Therefore: P₁V₁ = P₂V₂ We are given the initial pressure P₁ as 6.80 atm.
02

Calculate the pressure when the container is twice as large

We are told that the container becomes twice as large. Therefore, V₂ will be twice the initial volume. V₂ = 2V₁ Now, substitute this value of V₂ in the Boyle's Law equation: P₁V₁ = P₂(2V₁) Solve for P₂: P₂ = P₁V₁ / (2V₁) Since V₁ cancels out: P₂ = P₁ / 2 Now, substitute the given value of P₁: P₂ = 6.80 / 2 P₂ = 3.40 atm Hence, the pressure of CO₂ when the container is twice as large is 3.40 atm.
03

Calculate the pressure when the container is one-fourth as large

Now, we are told that the container becomes one-fourth as large. Therefore, V₂ will be one-fourth the initial volume. V₂ = V₁ / 4 Now, substitute this value of V₂ in the Boyle's Law equation: P₁V₁ = P₂(V₁ / 4) Solve for P₂: P₂ = P₁V₁ / (V₁ / 4) Since V₁ cancels out: P₂ = 4 * P₁ Now, substitute the given value of P₁: P₂ = 4 * 6.80 P₂ = 27.20 atm Hence, the pressure of CO₂ when the container is one-fourth as large is 27.20 atm.

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Most popular questions from this chapter

Explain why gases always occupy the entire container they are in, but solids and liquids do not.

To produce \(500.0 \mathrm{~g}\) of gaseous \(\mathrm{H}_{2} \mathrm{O}\), what volumes in liters of hydrogen gas and oxygen gas, both at STP, must combine via the reaction \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\)

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Consider a container that contains \(1.00\) mole of \(\mathrm{CO}_{2}(g)\) at \(298 \mathrm{~K}\). (a) What does the ideal gas law predict the pressure to be in atm? (b) What does the van der Waals equation predict the pressure to be? (c) What is the percent difference of the van der Waals pressure from the ideal pressure? (d) Suppose you increased the temperature to \(1000 \mathrm{~K}\). Would you expect the percent difference to increase or decrease compared to your answer in (c)? Explain.
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