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Ch.10 - Gases
Chapter 10, Problem 44d

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 210.0 L that contains O2 gas at a pressure of 16,500 kPa at 23 °C. (d) What would be the pressure of the gas, in kPa, if it were transferred to a container at 24 °C whose volume is 55.0 L?

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Step 1: We can use the ideal gas law to solve this problem. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin. However, since we are comparing the same gas under two different conditions, we can use the combined gas law which is (P1V1/T1) = (P2V2/T2).
Step 2: First, convert the temperatures from Celsius to Kelvin. The Kelvin temperature is the Celsius temperature plus 273.15. So, T1 = 23 °C + 273.15 = 296.15 K and T2 = 24 °C + 273.15 = 297.15 K.
Step 3: Substitute the known values into the combined gas law equation. P1 = 16,500 kPa, V1 = 210.0 L, T1 = 296.15 K, V2 = 55.0 L, and T2 = 297.15 K. We are solving for P2, the new pressure.
Step 4: Rearrange the equation to solve for P2. This gives P2 = (P1V1/T1) * (T2/V2).
Step 5: Plug in the known values and solve for P2. This will give you the new pressure of the gas in the smaller container at 24 °C.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin. This law allows us to predict how a gas will behave under different conditions.
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Charles's Law

Charles's Law states that the volume of a gas is directly proportional to its temperature when pressure is held constant. This means that as the temperature of a gas increases, its volume also increases, and vice versa. In the context of the question, this law helps to understand how temperature changes affect gas pressure when the volume is altered.
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Combined Gas Law

The Combined Gas Law combines Boyle's Law, Charles's Law, and Gay-Lussac's Law into a single equation, allowing for the calculation of the state of a gas when temperature, volume, and pressure change. It is expressed as (P1V1)/T1 = (P2V2)/T2. This law is particularly useful for solving problems where multiple variables change, as in the scenario presented in the question.
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Related Practice
Textbook Question

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 210.0 L that contains O2 gas at a pressure of 16,500 kPa at 23 °C. (a) What mass of O2 does the tank contain?

Textbook Question

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 210.0 L that contains O2 gas at a pressure of 16,500 kPa at 23 °C. (b) What volume would the gas occupy at STP?

Textbook Question

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 210.0 L that contains O2 gas at a pressure of 16,500 kPa at 23 °C. (c) At what temperature would the pressure in the tank equal 15.2 MPa?

Textbook Question

In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In 30 minutes the average cockroach (running at 0.08 km/h) consumed 1.0 mL of O2 at 101.33 kPa pressure and 20 °C per gram of insect mass. (a) How many moles of O2 would be consumed in 1 day by a 6.3-g cockroach moving at this speed?

Textbook Question

The physical fitness of athletes is measured by 'VO2 max,' which is the maximum volume of oxygen consumed by an individual during incremental exercise (for example, on a treadmill). An average male has a VO2 max of 45 mL O2/kg body mass/min, but a world-class male athlete can have a VO2 max reading of 88.0 mL O2/kg body mass/min. (a) Calculate the volume of oxygen, in mL, consumed in 1 hr by an average man who weighs 85 kg and has a VO2 max reading of 47.5 mL O2/kg body mass/min. (b) If this man lost 10 kg, exercised, and increased his VO2 max to 65.0 mL O2/kg body mass/min, how many mL of oxygen would he consume in 1 hr?

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