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Chapter 12: Problem 44

At what Kelvin temperature will \(37.5 \mathrm{~mol}\) of Ar occupy a volume of \(725 \mathrm{~L}\) at a pressure of 675 torr?

Short Answer

Expert verified
The temperature is approximately 208.85 K.

Step by step solution

01

Identify the Ideal Gas Law

The Ideal Gas Law equation is given by \[ PV = nRT \]where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin.
02

Convert Units if Necessary

Ensure all units are consistent with the gas constant used. The gas constant, R, is typically \(0.0821 \text{ L}·\text{atm}/\text{mol}·\text{K}\). Convert pressure from torr to atm using the conversion factor: \(1 \text{ atm} = 760 \text{ torr}\). Thus, \[ P = 675 \text{ torr} \times \left( \frac{1 \text{ atm}}{760 \text{ torr}} \right) = 0.888 \text{ atm} \]
03

Rearrange the Ideal Gas Law for Temperature

Rearrange the Ideal Gas Law equation to solve for temperature (T): \[ T = \frac{PV}{nR} \]
04

Substitute the Known Values

Substitute the values into the rearranged equation: \[ P = 0.888 \text{ atm}, V = 725 \text{ L}, n = 37.5 \text{ mol}, R = 0.0821 \text{ L}·\text{atm}/\text{mol}·\text{K} \]Thus,\[ T = \frac{0.888 \text{ atm} \times 725 \text{ L}}{37.5 \text{ mol} \times 0.0821 \text{ L}·\text{atm}/\text{mol}·\text{K}} \]
05

Calculate the Temperature

Perform the calculation:\[ T = \frac{643.2 \text{ atm·L}}{3.07875 \text{ atm·L/mol·K}} \approx 208.85 \text{ K} \]

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Key Concepts

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

Kelvin Temperature Calculation
To understand how to find the temperature of a gas using Kelvin, you first need to know what Kelvin is. Kelvin is a temperature scale where 0 Kelvin is absolute zero, the coldest possible temperature. It is the standard unit of temperature in scientific contexts. This problem is asking us to determine the temperature in Kelvin that will result when certain conditions are met for 37.5 moles of Argon gas.
Unit Conversion
Before solving for the temperature, we need to make sure all our units are correctly converted and consistent. In this case, pressure is originally given in torr but needs to be converted to atmospheres (atm) because the gas constant, R, we will use is reported in terms of atmospheres. The conversion factor is:
  • 1 atm = 760 torr
Therefore, we convert 675 torr to atmospheres as follows:
  • Pressure (P) = 675 torr × (1 atm / 760 torr) = 0.888 atm
Ideal Gas Law Equation
The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of gases. The equation is:
  • \[ PV = nRT \]
Where:
  • P is the pressure of the gas
  • V is the volume of the gas
  • n is the number of moles of the gas
  • R is the universal gas constant (0.0821 L·atm/mol·K)
  • T is the temperature in Kelvin
For our problem, we will rearrange this equation to solve for the temperature, T, using:\[ T = \frac{PV}{nR} \]
Pressure Conversion
Ensuring correct units for pressure is important in using the Ideal Gas Law. As mentioned earlier, our original pressure is given in torr, whereas the gas constant R we use is in atm. The conversion from torr to atm is crucial:
  • Pressure in atm = Pressure in torr × (1 atm / 760 torr)
After conversion, we found our pressure to be 0.888 atm.
Temperature Determination
Finally, we can determine the temperature using the converted units and the Ideal Gas Law. We input the known values into the rearranged equation:
  • \[ T = \frac{PV}{nR} \]
  • \[ P = 0.888 \text{ atm} \]
  • \[ V = 725 \text{ L} \]
  • \[ n = 37.5 \text{ mol} \]
  • \[ R = 0.0821 \text{ L}·\text{atm}/\text{mol}·\text{K} \]
Substituting these values, we get:\[ T = \frac{0.888 \text{ atm} \times 725 \text{ L}}{37.5 \text{ mol} \times 0.0821 \text{ L}·\text{atm}/\text{mol}·\text{K}} \approx 208.85 \text{ K} \]Thus, the temperature where 37.5 moles of Argon occupy 725 liters at 675 torr is approximately 208.85 Kelvin.

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Most popular questions from this chapter

$$ \begin{aligned} &\text { Fill in the table below with the missing information: }\\\ &\begin{array}{|c|c|c|c|} \hline & \text { torr } & \text { in. Hg } & \text { kilopascals (kPa) } \\ \hline \text { (a) } & & 30.2 & \\ \hline \text { (b) } & 752 & & \\ \hline \text { (c) } & & & 99.3 \\ \hline \end{array} \end{aligned} $$

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