What is the Celsius temperature of a gas if \(3.200\) moles of it occupy \(12.00 \mathrm{~L}\) at \(8.500 \mathrm{~atm} ?\)

Short Answer

Expert verified
The Celsius temperature of the gas is approximately 116.15°C.

Step by step solution

01

1. Write down known and unknown values

In this case, we have: P = 8.500 atm V = 12.00 L n = 3.200 moles R (gas constant) = 0.0821 L atm K⁻¹ mol⁻¹ [When pressure is in atm and volume is in L] T = ? (In Celsius)
02

2. Use the Ideal Gas Law to find the Kelvin temperature

We will first find the temperature in Kelvin using the Ideal Gas Law equation: PV = nRT. Rearrange the equation to solve for T: T = \(\frac{PV}{nR}\) Now, substitute the given values: T = \(\frac{(8.500 \text{ atm})(12.00 \text{ L})}{(3.200 \text{ moles})(0.0821 \text{ L atm K⁻¹ mol⁻¹})}\)
03

3. Calculate the Kelvin temperature

Multiply and divide the values to find the temperature in Kelvin: T = \(\frac{102.00 \text{ L atm}}{0.262 \text{ L atm K⁻¹ mol⁻¹} }\) T ≈ 389.3 K
04

4. Convert the Kelvin temperature to Celsius

Now, convert the temperature from Kelvin to Celsius using the formula: °C = K - 273.15 °C = 389.3 K - 273.15 °C ≈ 116.15
05

5. State the final answer

The Celsius temperature of the gas is approximately 116.15°C.

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Key Concepts

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

Celsius Temperature Conversion
Understanding how to convert temperature from Kelvin to Celsius is a fundamental skill in chemistry. Remember, when dealing with gases and the Ideal Gas Law, temperature must initially be in Kelvin to use the gas constant values correctly. However, to relate the findings to everyday situations or other scientific data, we often need the result in Celsius.

The conversion is straightforward: subtract 273.15 from the Kelvin temperature. In formula form, it looks like this: \[ ^\circ C = K - 273.15 \].

For example, converting a temperature of 389.3 K to Celsius involves calculating \( 389.3 K - 273.15 \), which equals approximately 116.15°C. This step is vital to present the information in a more commonly understood format.
PV=nRT
The Ideal Gas Law, represented as \( PV=nRT \), is a cornerstone equation in understanding the behavior of gases under various conditions of pressure (\( P \)), volume (\( V \)), moles of gas (\( n \)), and temperature (\( T \)). This relationship allows you to predict how a gas will respond when one or more of these variables are changed.

When given three of these variables, you can solve for the fourth by rearranging the equation appropriately. In the case of finding the temperature, you would isolate \( T \) to one side: \( T = \frac{PV}{nR} \). Following this step ensures you can solve for any missing variable with confidence.
Gas constant
The gas constant (\( R \)) is a crucial part of the Ideal Gas Law. Its value depends on the units of pressure, volume, and temperature. For most calculations, when pressure is measured in atmospheres (atm) and volume in liters (L), the gas constant used is \( R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1} \).

This constant provides the necessary link between the physical properties of the gas and allows the other variables to be related to each other predictably. Always ensure that the units you're using for pressure and volume match the units of the gas constant.
Moles and Gas Volume Relationship
The relation between moles (\( n \)) and volume (\( V \)) of a gas is one of direct proportionality, as evidenced by the Ideal Gas Law. Keeping temperature and pressure constant, an increase in the number of moles of gas will lead to a proportional increase in volume, and vice versa.

This aspect of the Ideal Gas Law means that we can predict how the volume of gas changes with the amount of substance present. It also underlies concepts such as molar volume at STP (standard temperature and pressure), where 1 mole of any gas occupies 22.4 liters.

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Most popular questions from this chapter

In the Haber process, nitrogen reacts with hydrogen to produce ammonia: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)\) (a) Suppose \(2.0 \mathrm{~L}\) of \(\mathrm{N}_{2}\) gas at STP is combined with \(6.0 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas, with the two gases being at the same temperature and pressure. Is this reaction being run in a balanced fashion or in a limiting fashion? Explain how you can tell without doing any calculations. (b) If \(50.0 \mathrm{~L}\) of \(\mathrm{N}_{2}\) gas at \(200.0 \mathrm{lb} / \mathrm{in}^{2}\) and \(22.0^{\circ} \mathrm{C}\) is combined with \(100.0 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas at \(240.0 \mathrm{lb} / \mathrm{in}^{2}\) and \(22.0^{\circ} \mathrm{C}\), what mass in grams of ammonia is produced? \(\left[14.70 \mathrm{lb} / \mathrm{in}^{2}=760.0 \mathrm{~mm} \mathrm{Hg}\right]\)

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