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An ideal gas at \(15.5^{\circ} \mathrm{C}\) and a pressure of \(1.72 \times 10^{5}\) Pa occupies a volume of 2.81 \(\mathrm{m}^{3} .\) (a) How many moles of gas are present? (b) If the volume is raised to 4.16 \(\mathrm{m}^{3}\) and the temperature raised to \(28.2^{\circ} \mathrm{C},\) what will be the pressure of the gas?

Short Answer

Expert verified
(a) 20.2 moles. (b) 鈮 1.215 脳 10鈦 Pa.

Step by step solution

01

Convert Temperature to Kelvin

First, we need to convert the given temperatures from degrees Celsius to Kelvin. The conversion formula is: \[ T(K) = T(^{\circ}C) + 273.15 \]Initial Temperature: \[ T_1 = 15.5 + 273.15 = 288.65\, K \]Final Temperature: \[ T_2 = 28.2 + 273.15 = 301.35\, K \]
02

Use Ideal Gas Law to Find Moles

To find the number of moles \( n \), use the ideal gas law: \[ PV = nRT \]Where:- \( P = 1.72 \times 10^5 \) Pa- \( V = 2.81 \) m鲁- \( R = 8.314 \) J/mol路K (ideal gas constant)- \( T = 288.65 \) KRearrange for \( n \):\[ n = \frac{PV}{RT} = \frac{1.72 \times 10^5 \times 2.81}{8.314 \times 288.65} \approx 20.2 \] moles
03

Use Ideal Gas Law for Final Conditions

With the number of moles determined, use the ideal gas law again for the new conditions to find the final pressure \( P_2 \): First, recall the rearranged ideal gas law: \[ P_2 V_2 = nRT_2 \]We know:- \( n \approx 20.2 \) moles- \( V_2 = 4.16 \) m鲁- \( T_2 = 301.35 \) K- \( R = 8.314 \) J/mol路KSolve for \( P_2 \):\[ P_2 = \frac{nRT_2}{V_2} = \frac{20.2 \times 8.314 \times 301.35}{4.16} \approx 1.215 \times 10^5 \] Pa
04

Communicate the Final Answer

(a) The number of moles of gas present is approximately 20.2 moles.(b) The final pressure of the gas after the change in volume and temperature is approximately \( 1.215 \times 10^5 \) Pa.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature Conversion
When dealing with gases, it's important to use the correct temperature scale. For gas laws, like the Ideal Gas Law, temperatures must be in Kelvin. This is because Kelvin is an absolute temperature scale, which starts at absolute zero鈥攖he point where theoretically, there is no molecular motion. The formula to convert Celsius to Kelvin is simple:
  • \[ T(K) = T(^{\circ}C) + 273.15 \]
To use this for our exercise, just add 273.15 to the Celsius temperature. It's always a good first step in these kinds of problems. Remember, wrong temperature conversions can lead to errors in your final calculations, as all subsequent calculations depend on accurate initial conditions.
The initial temperature of 15.5 掳C becomes 288.65 K, and the final temperature of 28.2 掳C becomes 301.35 K after conversion.
Pressure Change
Pressure change is an important concept in problems involving gases. As conditions like volume and temperature change, so too does the pressure, assuming the amount of gas remains constant. This relationship is part of the Ideal Gas Law, which states that the pressure of a gas is proportional to the temperature and quantity, and inversely proportional to the volume.
In our problem, starting from an initial pressure of \(1.72 \times 10^{5}\) Pa, we calculated a new pressure of approximately \(1.215 \times 10^{5}\) Pa when both temperature and volume were changed. Notice how these changes ultimately influenced the drop in pressure. It's also a good practice to ensure that pressure units remain consistent, typically in Pascals (Pa), when using the Ideal Gas Law.
Gas Moles Calculation
Calculating the number of moles in a gas sample is fundamental when using the Ideal Gas Law:
  • \[ PV = nRT \]
This equation allows us to find the exact amount of gas when pressure (\( P \)), volume (\( V \)), and temperature (\( T \)) are known.
Rearranging to solve for moles (\( n \)) gives:
  • \[ n = \frac{PV}{RT} \]
In our example, with an initial pressure of \(1.72 \times 10^{5} \) Pa, volume of 2.81 m鲁 and Kelvin temperature, we find about 20.2 moles of gas. The concepts here highlight that even minor changes in any variable might significantly affect the outcome, due to their direct proportional relationship.
Volume Change
Volume changes have a big impact on gas behavior. As per the Ideal Gas Law, as the volume occupied by the gas increases at constant temperature, the pressure decreases if the amount of moles remains the same.
For our gas, the volume was increased from 2.81 m鲁 to 4.16 m鲁. To find the new pressure, the Ideal Gas Law was used while keeping the moles constant. This significant change in volume shows us how gases will expand to fill a larger space if they can, resulting in a lower pressure, confirming the inversely proportional nature of volume and pressure in gases.
Ideal Gas Constant
A pivotal part of the Ideal Gas Law is the Ideal Gas Constant (\( R \)), which bridges the units across energy, volume, and pressure. The constant \( R \) has a value of 8.314 J/mol路K.
Think of \( R \) as a conversion factor, ensuring the different units in the Ideal Gas Law formula balance out correctly to provide meaningful results.
  • It's important because:
    • It ensures that when we multiply the other terms (number of moles, temperature, and sometimes volume), they give us energy-equivalent units in joules.
Using \( R \) correctly is crucial to solving problems accurately, which ensures your final answers are in the correct context, whether that's in units of pressure, volume, or moles.

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Most popular questions from this chapter

Two moles of an ideal gas are placed in a container whose volume is \(8.5 \times 10^{-3} \mathrm{m}^{3} .\) The absolute pressure of the gas is \(4.5 \times 10^{5} \mathrm{Pa}\) . What is the average translational kinetic energy of a molecule of the gas?

A tank contains 0.85 mol of molecular nitrogen \(\left(\mathrm{N}_{2}\right)\) Determine the mass (in grams) of nitrogen that must be removed from the tank in order to lower the pressure from 38 to 25 atm. Assume that the volume and temperature of the nitrogen in the tank do not change.

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A certain element has a mass per mole of 196.967 g/mol. What is the mass of a single atom in (a) atomic mass units and (b) kilograms? (c) How many moles of atoms are in a 285 -g sample?

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