Calculating Final Temperature of a Perfect Gas at Constant Pressure

The question asks for the final temperature of a perfect gas when it is heated at constant pressure, causing its volume to double from an initial temperature of 25°C.

This scenario describes an isobaric process for a perfect gas. For a fixed amount of gas, the relationship between volume and temperature at constant pressure is governed by Charles's Law, which is a specific case of the Ideal Gas Law.

The Ideal Gas Law is given by:

\(PV = nRT\)

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 ideal gas constant.
• \(T\) is the absolute temperature of the gas (in Kelvin).

In this problem, the pressure \(P\) and the amount of gas \(n\) (and \(R\)) are constant. Therefore, the equation simplifies to:

\(V/T = \text{constant}\)

This means that for an isobaric process, the ratio of volume to absolute temperature is constant. We can write this relationship for the initial and final states of the gas:

\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)

Where:

• \(V_1\) is the initial volume.
• \(T_1\) is the initial absolute temperature.
• \(V_2\) is the final volume.
• \(T_2\) is the final absolute temperature.

We are given the initial temperature \(T_1 = 25^\circ \text{C}\). To use Charles's Law, we must convert this temperature to the absolute Kelvin scale. The conversion formula is:

\(T(\text{K}) = T(^\circ \text{C}) + 273.15\)

Using a rounded value of 273 for the conversion:

\(T_1 = 25^\circ \text{C} + 273 = 298 \text{ K}\)

We are also told that the final volume \(V_2\) is double the initial volume \(V_1\):

\(V_2 = 2V_1\)

Now, substitute these values into Charles's Law equation:

\(\frac{V_1}{T_1} = \frac{2V_1}{T_2}\)

We can cancel \(V_1\) from both sides (assuming \(V_1 \ne 0\)):

\(\frac{1}{T_1} = \frac{2}{T_2}\)

Now, solve for \(T_2\):

\(T_2 = 2 \times T_1\)

Substitute the value of \(T_1\) in Kelvin:

\(T_2 = 2 \times 298 \text{ K}\)

\(T_2 = 596 \text{ K}\)

The question asks for the final temperature in degrees Celsius. Convert \(T_2\) back to Celsius:

\(T_2(^\circ \text{C}) = T_2(\text{K}) - 273.15\)

Using the rounded value of 273:

\(T_2(^\circ \text{C}) = 596 \text{ K} - 273\)

\(T_2(^\circ \text{C}) = 323^\circ \text{C}\)

So, the final temperature when the volume is doubled at constant pressure is 323°C.

Step-by-Step Solution

1. Identify the initial temperature and the process type (constant pressure).
2. Note the relationship between initial and final volumes.
3. Convert the initial temperature from Celsius to Kelvin: \(T_1 = 25^\circ \text{C} + 273 = 298 \text{ K}\).
4. Apply Charles's Law for constant pressure process: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\).
5. Substitute \(V_2 = 2V_1\) into the equation: \(\frac{V_1}{T_1} = \frac{2V_1}{T_2}\).
6. Solve for the final absolute temperature \(T_2\): \(T_2 = 2T_1 = 2 \times 298 \text{ K} = 596 \text{ K}\).
7. Convert the final temperature back to Celsius: \(T_2 = 596 \text{ K} - 273 = 323^\circ \text{C}\).

Relevant Gas Laws and Processes

Understanding different gas processes is key to solving such problems:

• Isobaric Process: Occurs at constant pressure (\(P\)). Charles's Law (\(V \propto T\)) applies.
• Isochoric Process: Occurs at constant volume (\(V\)). Gay-Lussac's Law (\(P \propto T\)) applies.
• Isothermal Process: Occurs at constant temperature (\(T\)). Boyle's Law (\(P \propto 1/V\)) applies.
• Adiabatic Process: Occurs without heat transfer (\(Q=0\)). The relationship is \(PV^\gamma = \text{constant}\) (where \(\gamma\) is the adiabatic index).

In this specific question, the process is isobaric because the gas is heated at constant pressure.

Revision Table: Perfect Gas Laws

Additional Information on Temperature Scales

Temperature scales are crucial in gas laws because the relationships often depend on absolute temperature. The two main scales are Celsius and Kelvin.

• Celsius scale (°C): A relative scale where 0°C is the freezing point of water and 100°C is the boiling point of water at standard atmospheric pressure.
• Kelvin scale (K): An absolute scale where 0 K (absolute zero) is the theoretical point at which particles have minimum motion. It is related to Celsius by \(T(\text{K}) = T(^\circ \text{C}) + 273.15\). Gas law calculations involving \(V/T\) or \(P/T\) must use Kelvin.

Using the Celsius scale directly in ratios like \(V_1/T_1 = V_2/T_2\) would lead to incorrect results because the Celsius scale does not start at absolute zero.