Problem 48
Question
A student thinks he remembers reading that if you double the temperature of an ideal gas, its. pressure doubles. He is given a problem where he has an ideal gas at \(25.0{ }^{\circ} \mathrm{C}\) and \(2.5 \mathrm{~atm}\). He is asked what the temperature must be raised to in order to double the pressure to \(5.0 \mathrm{~atm}\). He answers, \(^{\prime \prime} 50.0{ }^{\circ} \mathrm{C}\), of course." Why is he wrong? What lesson should he learn about using the ideal gas law? What is the temperature increase in Celsius degrees that will double the pressure?
Step-by-Step Solution
Verified Answer
The student is wrong because they didn't use the ideal gas law and express the temperature in Kelvin. The main lesson to be learned here is to use the ideal gas law correctly and always express temperature in Kelvin. The temperature increase in Celsius degrees required to double the pressure is 298.15 °C.
1Step 1: Convert initial temperature to Kelvin
First, convert the given initial temperature from Celsius to Kelvin. The conversion formula is T(K) = T(°C) + 273.15, where T(K) is the temperature in Kelvin and T(°C) is the temperature in Celsius.
T(K) = 25.0 + 273.15 = 298.15 K
2Step 2: Apply the ideal gas law
The ideal gas law states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Since the volume and the number of moles does not change, we can simplify the equation as follows:
\(nR\left(\frac{P_1}{T_1}\right)=nR\left(\frac{P_2}{T_2}\right)\)
Given that the pressure is doubled, \(P_2\) = 2 * \(P_1\). Thus, we can cancel out nR from both sides and rearrange the equation to solve for the final temperature, \(T_2\):
\(T_2 = \frac{P_2}{P_1} \cdot T_1\)
3Step 3: Calculate the final temperature in Kelvin
Now, plug in the values for the pressure ratio and initial temperature, and calculate the final temperature in Kelvin:
\(T_2 = \frac{5.0 \mathrm{~atm}}{2.5 \mathrm{~atm}} \cdot 298.15 K = 596.30 K\)
4Step 4: Convert the final temperature to Celsius
Finally, convert the final temperature from Kelvin to Celsius using the conversion formula:
T(°C) = T(K) - 273.15
T(°C) = 596.30 K - 273.15 = 323.15 °C
To determine the temperature increase in Celsius degrees, subtract the initial temperature from the final temperature:
Temperature increase = 323.15 °C - 25.0 °C = 298.15 °C
The student is wrong because they didn't use the ideal gas law and express the temperature in Kelvin. The temperature increase in Celsius degrees required to double the pressure is 298.15 °C.
Key Concepts
Temperature-Pressure RelationshipKelvin Temperature ScaleConverting Celsius to KelvinPV=nRT Equation
Temperature-Pressure Relationship
Understanding the relationship between temperature and pressure is crucial when studying gases. According to the ideal gas law, for a given amount of gas at a constant volume, an increase in temperature will result in an increase in pressure. This is because temperature is a measure of the average kinetic energy of the gas particles; as temperature rises, particles move faster and collide with the walls of their container more forcefully, thereby increasing the pressure.
It's important to note that this relationship assumes all other factors remain constant, and it's particularly true when temperatures are expressed in Kelvin, as the Kelvin scale is directly proportional to the kinetic energy of the particles. The student in the exercise mistakenly assumed that doubling the temperature in degrees Celsius would double the pressure, ignoring that temperature in the ideal gas law must be in Kelvin for it to be directly proportional to pressure.
It's important to note that this relationship assumes all other factors remain constant, and it's particularly true when temperatures are expressed in Kelvin, as the Kelvin scale is directly proportional to the kinetic energy of the particles. The student in the exercise mistakenly assumed that doubling the temperature in degrees Celsius would double the pressure, ignoring that temperature in the ideal gas law must be in Kelvin for it to be directly proportional to pressure.
Kelvin Temperature Scale
The Kelvin temperature scale is fundamental in scientific equations, including the ideal gas law, because it starts at absolute zero—the point at which particles have minimal motion. Unlike Celsius or Fahrenheit, Kelvin is not referred to in degrees; it is an absolute temperature scale. This makes the Kelvin scale the preferred choice in scientific computations that involve temperature. One key factor to always remember is that Kelvin allows for direct proportionality in calculations, which means that when you double the Kelvin temperature, you're also doubling the kinetic energy of the particles.
Converting Celsius to Kelvin
Converting Celsius to Kelvin is a simple yet critical step in problems involving the ideal gas law. To convert, add 273.15 to the Celsius temperature. This conversion aligns the temperature with the absolute scale used in the ideal gas law, where absolute zero (0 K) is the base point.
For instance, 25.0 °C equals 298.15 K. This conversion ensures that calculations involving temperature changes reflect changes in kinetic energy accurately. It's important to avoid a common misconception that a temperature doubling in Celsius will result in a doubling in Kelvin; because of the 273.15 constant difference, this is not the case.
For instance, 25.0 °C equals 298.15 K. This conversion ensures that calculations involving temperature changes reflect changes in kinetic energy accurately. It's important to avoid a common misconception that a temperature doubling in Celsius will result in a doubling in Kelvin; because of the 273.15 constant difference, this is not the case.
PV=nRT Equation
The PV=nRT equation, known as the ideal gas law, is a cornerstone of thermodynamics. It connects pressure (P), volume (V), the number of moles (n), the ideal gas constant (R), and temperature in Kelvin (T) in a single concise relation. Specifically, it states that the product of pressure and volume of a gas equals the product of the number of moles, the ideal gas constant, and the absolute temperature.
In the context of the exercise, the student mishandled this equation by not considering the Kelvin temperature. Proper use of the ideal gas law would have shown that to double the pressure of a gas at constant volume and moles, the absolute temperature must also be doubled. This equation's beauty lies in its ability to predict how a gas will behave under various conditions, assuming the gas is ideal and that all variables are properly understood and utilized.
In the context of the exercise, the student mishandled this equation by not considering the Kelvin temperature. Proper use of the ideal gas law would have shown that to double the pressure of a gas at constant volume and moles, the absolute temperature must also be doubled. This equation's beauty lies in its ability to predict how a gas will behave under various conditions, assuming the gas is ideal and that all variables are properly understood and utilized.
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