Question

A gas at \(27^{\circ} \mathrm{C}\) temperature and 30 atmospheric pressure \(1 \mathrm{~s}\) allowed to expand to the atmospheric pressure if the volume becomes two times its initial volume, then the final temperature becomes (B) \(-173^{\circ} \mathrm{C}\) (A) \(273^{\circ} \mathrm{C}\) (C) \(173^{\circ} \mathrm{C}\) (D) \(100^{\circ} \mathrm{C}\)

Short answer

The final temperature after the gas expands to atmospheric pressure and doubles its initial volume is approximately \(-253.14^{\circ}\mathrm{C}\). None of the given options match this solution, but option (B) is the closest. There might be an error in the options provided.

Step-by-step solution

Write down the Combined Gas Law formula

First, we need to write down the formula for the Combined Gas Law, which is: \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\] Here, \(P_1\) and \(P_2\) are the initial and final pressures, \(V_1\) and \(V_2\) are the initial and final volumes, and \(T_1\) and \(T_2\) are the initial and final temperatures of the gas.

Convert temperature to Kelvin

To solve this problem, we need to ensure that all temperatures are in Kelvin (K), because gas laws only apply to temperatures in K. So, we need to convert the given temperature of \(27^{\circ}\mathrm{C}\) to Kelvin: \[T_1 = 27 + 273.15 = 300.15\,\mathrm{K}\]

Convert pressure to atmospheres

The problem states that the initial pressure is 30 atmospheric pressures. As we also have the final pressure being 1 atmospheric pressure, it means that our pressures are already in the correct units.

Substitute the given values into the Combined Gas Law formula

Now, we will substitute the given values into the formula and solve for the final temperature, \(T_2\). Since the final volume is two times the initial volume, we have \(V_2 = 2V_1\). Plug this into the Combined Gas Law formula and rearrange: \[\frac{P_1 V_1}{T_1} = \frac{P_2 (2V_1)}{T_2}\] Now, we can substitute the given values: \[\frac{30\,\cancel{V_1}}{300.15} = \frac{1\,\cancel{V_1} \cdot 2}{T_2}\]

Solve for the final temperature, \(T_2\)

Now, solve for \(T_2\): \[\frac{30}{300.15} = \frac{2}{T_2}\] \[\Rightarrow T_2 = \frac{2\cdot 300.15}{30} = \frac{600.3}{30} = 20.01\, \mathrm{K}\]

Convert final temperature back to Celsius

Finally, convert the final temperature back to Celsius: \[T_2 = 20.01 - 273.15 = -253.14^{\circ} \mathrm{C}\] Here, we observe that none of the given options match our solution. However, option (B) seems to be the closest to the solution. This might be an error in the options provided, and you may want to double-check the options with your teacher.

Key concepts

Temperature Conversion

When dealing with gas laws, converting temperature is crucial. These laws require temperatures to be in Kelvin. Kelvin is the SI unit of temperature, and it is used as it starts from absolute zero—the point where particles have minimum thermal energy. The conversion from Celsius to Kelvin is simple: you just add 273.15 to the Celsius temperature.

For example, in the exercise, the initial temperature given is \(27^{\circ}\mathrm{C}\). To convert this to Kelvin:

• Add 273.15 to 27.
• This gives \(T_1 = 300.15\,\mathrm{K}\).

This conversion ensures you're working within the appropriate framework for solving gas law equations.

Gas Expansion

Gas expansion occurs when a gas is allowed to occupy a larger volume, often as the pressure or temperature conditions change. In the context of the Combined Gas Law, this situation maintains a connection between pressure, volume, and temperature.

For the problem at hand, the gas expands until its volume is double the original volume. This doubling means in mathematical terms, \(V_2 = 2V_1\).

This is key because according to the Combined Gas Law, the relationships between these three variables - pressure, volume, and temperature - are interconnected. When you understand how one influences the other, it becomes easier to predict the behavior of gases.

Pressure-Volume Relationship

Understanding the pressure-volume relationship is essential when working with the Combined Gas Law. According to Boyle's Law, which is a part of the Combined Gas Law, the volume of a gas is inversely proportional to its pressure when the temperature is held constant. However, when temperature also changes, this relationship becomes more complex.

In the exercise, the initial pressure \(P_1\) is 30 atm, and by the time the gas expands, the pressure drops to atmospheric pressure, \(P_2 = 1\,\mathrm{atm}\).

• As volume increases, pressure decreases if the gas expands freely.
• Using the Combined Gas Law, the variables can be used to find the unknown temperature \(T_2\) after expansion.

Remember, this relationship demonstrates how pressure and volume are loads that "see-saw" based on temperature changes, encapsulating the very dynamic nature of gases.