To determine the scenario in which the volume of an enclosed gas remains constant, we can use the Ideal Gas Law given by:
\[ PV = nRT \]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
For the volume to remain constant, the ratio of pressure to temperature must also remain constant. This can be expressed as:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Let's evaluate each of the scenarios:
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Tripling the pressure and taking one-third of the temperature: \[ P_2 = 3P_1, \quad T_2 = \frac{1}{3}T_1 \] \[ \frac{P_2}{T_2} = \frac{3P_1}{\frac{1}{3}T_1} = 9 \frac{P_1}{T_1} \] This changes the ratio, so the volume would not remain constant.
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Doubling the pressure and quadrupling the temperature: \[ P_2 = 2P_1, \quad T_2 = 4T_1 \] \[ \frac{P_2}{T_2} = \frac{2P_1}{4T_1} = \frac{1}{2} \frac{P_1}{T_1} \] This changes the ratio, so the volume would not remain constant.
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Doubling the pressure and taking one-half of the temperature: \[ P_2 = 2P_1, \quad T_2 = \frac{1}{2}T_1 \] \[ \frac{P_2}{T_2} = \frac{2P_1}{\frac{1}{2}T_1} = 4 \frac{P_1}{T_1} \] This changes the ratio, so the volume would not remain constant.
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Tripling the pressure and also tripling the temperature: \[ P_2 = 3P_1, \quad T_2 = 3T_1 \] \[ \frac{P_2}{T_2} = \frac{3P_1}{3T_1} = \frac{P_1}{T_1} \] This keeps the ratio the same, so the volume would remain constant.
Therefore, the scenario in which the volume of an enclosed gas remains constant is tripling the pressure of the gas and also tripling the temperature, in Kelvin, of the gas.