A container of fixed volume contains a gas at 27°C. To double the pressure of the gas, the temperature of the gas should be raised to °C.
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Correct Answer: 327
Approach Solution - 1
To solve the problem of determining the temperature needed to double the pressure of a gas within a fixed volume, we apply the Ideal Gas Law in the form of Charles's law, which is given by:
\( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)
Here, \(P_1\) is the initial pressure, \(T_1\) is the initial temperature in Kelvin, \(P_2\) is the final pressure, and \(T_2\) is the final temperature in Kelvin.
The initial condition is at 27°C, which is 300K (since \(T(K) = T(°C) + 273\)). We aim to double the pressure (\(P_2 = 2P_1\)).
Solving for \(T_2\):
\[ \frac{P_1}{300} = \frac{2P_1}{T_2} \]
By simplifying, we find:
\[ T_2 = 2 \times 300 = 600 \, \text{K} \]
Convert back to Celsius: \(T(°C) = T(K) - 273\):
\(T_2 = 600 - 273 = 327°C\).
Thus, the temperature should be raised to 327°C to double the pressure.
Approach Solution -2
We are given that a gas is contained in a fixed-volume container at an initial temperature of 27°C. We need to find the temperature (in °C) to which the gas should be heated so that its pressure becomes double.
Step 2: Use the gas law relation for constant volume.
For a fixed mass of gas at constant volume, the pressure is directly proportional to the absolute temperature:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \] where \( P_1 \) and \( P_2 \) are the initial and final pressures, and \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.
Step 3: Substitute given values.
We are told that the final pressure is double the initial pressure, so \( P_2 = 2P_1 \).
Hence, \[ \frac{P_1}{T_1} = \frac{2P_1}{T_2} \] Simplifying, we get: \[ T_2 = 2T_1 \]
Step 4: Convert temperature to Kelvin and compute.
The initial temperature is 27°C, which in Kelvin is: \[ T_1 = 27 + 273 = 300 \, K \] Therefore, \[ T_2 = 2 \times 300 = 600 \, K \] Now, converting back to Celsius: \[ T_2 = 600 - 273 = 327°C \]
Final Answer:
\[ \boxed{327^\circ \text{C}} \]
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