Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4% when heated by \( 1^\circ \text{C} \). Its initial temperature must be :
Show Hint
- \( 25^\circ \text{C} \)
- \( 2500 \, \text{K} \)
- \( 250 \, \text{K} \)
- \( 250^\circ \text{C} \)
The Correct Option is C
Approach Solution - 1
To solve this problem, we will use the ideal gas law, specifically the relationship between pressure, temperature, and volume. The problem states that pressure increases by 0.4% when the temperature is increased by \( 1^\circ \text{C} \), and we need to find the initial temperature.
The ideal gas law is given by:
\(PV = nRT\)
Here, \(P\) is pressure, \(V\) is volume, \(n\) is number of moles, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.
Since the vessel is closed, volume \(V\) and number of moles \(n\) remain constant. Therefore, the relation between pressure and temperature is:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)
Where:
- \(P_1\) = initial pressure
- \(T_1\) = initial temperature
- \(P_2\) = final pressure
- \(T_2\) = final temperature
We are given that pressure increases by 0.4%, so
\(P_2 = P_1 \times 1.004\)
And the temperature is increased by \(1^\circ \text{C}\), so the final temperature \(T_2\) is:
\(T_2 = T_1 + 1\)
Substitute these into the equation:
\(\frac{P_1}{T_1} = \frac{P_1 \times 1.004}{T_1 + 1}\)
Cancelling out \(P_1\) from both sides, we get:
\(\frac{1}{T_1} = \frac{1.004}{T_1 + 1}\)
Cross-multiplying gives:
\(T_1 + 1 = T_1 \times 1.004\)
Rearranging the terms:
\(1 = 0.004 \times T_1\)
Thus:
\(T_1 = \frac{1}{0.004}\)
\(T_1 = 250 \, \text{K}\)
Therefore, the initial temperature of the gas was \( 250 \, \text{K} \). Hence, the correct answer is: \( 250 \, \text{K} \)
Approach Solution -2
To determine the initial temperature of an ideal gas in a closed vessel where the pressure increases by 0.4% upon heating by \(1^\circ \text{C}\), we use the Ideal Gas Law and the concept of pressure change with temperature.
The Ideal Gas Law is given by:
\(PV = nRT\)
Where:
- \(P\) is the pressure
- \(V\) is the volume
- \(n\) is the number of moles
- \(R\) is the universal gas constant
- \(T\) is the temperature in Kelvin
Since the vessel is closed, the volume \(V\) and number of moles \(n\) remain constant. As a result, the relationship between pressure and temperature for given changes is:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)
Let:
- \(P_1\) be the initial pressure,
- \(T_1\) be the initial temperature in Kelvin,
- \(P_2 = P_1(1 + 0.004)\) (since the pressure increased by 0.4%),
- \(T_2 = T_1 + 1\) (due to a temperature increase of \(1^\circ \text{C}\)).
Substituting these into the equation gives:
\(\frac{P_1}{T_1} = \frac{P_1(1 + 0.004)}{T_1 + 1}\)
Cancel \(P_1\) from both sides:
\(\frac{1}{T_1} = \frac{1 + 0.004}{T_1 + 1}\)
Cross-multiply to get:
\(T_1 + 1 = T_1(1 + 0.004)\)
Expanding and solving for \(T_1\):
\(T_1 + 1 = T_1 + 0.004T_1\) \(1 = 0.004T_1\) \(T_1 = \frac{1}{0.004} = 250 \, \text{K}\)
Therefore, the initial temperature of the gas is \(250 \, \text{K}\).
Thus, the correct option is: \( 250 \, \text{K} \)
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