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Question

A perfect gas at 27°C is heated at constant pressure till its volume is doubled. The final temperature is

The correct answer is

327°C

Perfect Gas Temperature Change

This problem involves understanding the relationship between the volume and temperature of a perfect gas when heated at a constant pressure. This scenario is perfectly described by Charles's Law, a fundamental principle in gas dynamics.

Gas Problem Analysis

Let's first identify the given information and what we need to determine in this perfect gas problem:

  • Initial Temperature (\(T_1\)): The gas starts at \(27^\circ\text{C}\). For gas law calculations, temperatures must always be converted to the absolute Kelvin scale.
  • Process Condition: The gas is heated at constant pressure. This is a crucial detail that points us to a specific gas law.
  • Volume Change: The volume of the gas is doubled. If the initial volume is \(V_1\), then the final volume \(V_2\) will be \(2V_1\).
  • Goal: We need to find the final temperature (\(T_2\)) of the gas in degrees Celsius.

Charles's Law Application

Since the heating process occurs at a constant pressure, Charles's Law is the appropriate gas law to use. Charles's Law states that for a fixed amount of gas at constant pressure, the volume is directly proportional to its absolute temperature.

Mathematically, Charles's Law can be expressed as:

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

Where:

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

Temperature Calculation Steps

Let's proceed with the step-by-step calculation to find the final temperature.

  1. Convert Initial Temperature to Kelvin:

    The initial temperature \(T_1 = 27^\circ\text{C}\) must be converted to Kelvin by adding \(273\) (or \(273.15\) for higher precision, but \(273\) is commonly used in such problems).

    \( T_1 = 27^\circ\text{C} + 273 = 300 \text{ K} \)

  2. Set Up Charles's Law Equation:

    We are given that the final volume \(V_2\) is double the initial volume \(V_1\), so \(V_2 = 2V_1\). Now, substitute the known values into Charles's Law equation:

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

    \( \frac{V_1}{300 \text{ K}} = \frac{2V_1}{T_2} \)

  3. Solve for the Final Temperature (\(T_2\)) in Kelvin:

    We can cancel \(V_1\) from both sides of the equation, as it appears on both numerator sides:

    \( \frac{1}{300 \text{ K}} = \frac{2}{T_2} \)

    Now, cross-multiply to solve for \(T_2\):

    \( T_2 = 2 \times 300 \text{ K} \)

    \( T_2 = 600 \text{ K} \)

  4. Convert Final Temperature Back to Celsius:

    The options are given in degrees Celsius, so we convert the final temperature from Kelvin back to Celsius.

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

    \( T_2 (\text{in }^\circ\text{C}) = 600 - 273 \)

    \( T_2 (\text{in }^\circ\text{C}) = 327^\circ\text{C} \)

Final Temperature Result

The final temperature of the perfect gas after being heated at constant pressure until its volume doubles is 327°C.

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Important Questions from Thermodynamics System and Processes

  1. Select the right expression for Vander Waal’s equation for a real gas.

  2. Which of the following pairs of thermodynamic properties can uniquely fix the state of a substance like water, irrespective of its thermodynamic state?

  3. A Monoatomic ideal gas (y = 1.67 Molecular weight = 40) is compressed adiabatically from 0.1 Mpa 300K to 0.2Mpa. The universal gas constant is 8.314KJ/kgK. The work of the compression of gas in KJ/kg is. (Take (2)y-1/y = 1.32)

  4. A reversible polytropic process can be described by

  5. A gas goes through a process given by pv2 = c. It expands from state 1 of 350 kPa and 0.049 m3 to a final pressure of 150 kPa. Work delivered during this process in kJ is:

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