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Chapter 5: Problem 56

A \(2.50-\mathrm{L}\) flask contains 0.60 \(\mathrm{g} \mathrm{O}_{2}\) at a temperature of \(22^{\circ} \mathrm{C}\) . What is the pressure (in atm) inside the flask?

Short Answer

Expert verified
The pressure inside the flask is approximately \(0.184\,\text{atm}\).

Step by step solution

01

1. Convert the temperature from Celsius to Kelvin

To convert the temperature from Celsius to Kelvin: \(T_K = T_C + 273.15\) In this case, the transformation will be: \(22^{\circ}\mathrm{C} + 273.15\) K. So, \(T_K = 295.15\,\text{K}\).
02

2. Convert the given mass of gas to moles

To find the number of moles, we'll use the molecular weight of oxygen, O2 (32 g/mol): \(n = \frac{\text{mass}}{\text{molecular weight}}\) \(n = \frac{0.60\,\text{g}}{32\,\frac{\text{g}}{\text{mol}}}\) \(n \approx 0.01875\,\text{mol}\)
03

3. Calculate the pressure using the Ideal Gas Law

Now that we have the number of moles of gas (\(n \approx 0.01875\,\text{mol}\)), the temperature in Kelvin (\(T_K = 295.15\,\text{K}\)), and the flask volume (\(V = 2.50\,\text{L}\)), we can calculate the pressure using the Ideal Gas Law formula: \(P=\frac{nRT}{V}\) \(P=\frac{(0.01875\,\text{mol})(0.0821\,\frac{\text{L}\cdot\text{atm}}{\text{K}\cdot\text{mol}})(295.15\,\text{K})}{2.50\,\text{L}}\) \(P \approx 0.184\,\text{atm}\) So, the pressure inside the flask is approximately \(0.184\,\text{atm}\).

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Most popular questions from this chapter

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