A perfect gas at 27°C is heated at constant pressure till its volume is doubled. The final temperature is
327°C
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.
Let's first identify the given information and what we need to determine in this perfect gas problem:
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:
Let's proceed with the step-by-step calculation to find the final temperature.
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).
\( T_1 = 27^\circ\text{C} + 273 = 300 \text{ K} \)
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:
\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \)
\( \frac{V_1}{300 \text{ K}} = \frac{2V_1}{T_2} \)
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 \(T_2\):
\( T_2 = 2 \times 300 \text{ K} \)
\( T_2 = 600 \text{ K} \)
The options are given in degrees Celsius, so we convert the final temperature from Kelvin back to Celsius.
\( T_2 (\text{in }^\circ\text{C}) = T_2 (\text{in K}) - 273 \)
\( T_2 (\text{in }^\circ\text{C}) = 600 - 273 \)
\( T_2 (\text{in }^\circ\text{C}) = 327^\circ\text{C} \)
The final temperature of the perfect gas after being heated at constant pressure until its volume doubles is 327°C.
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