In the scenario you described, we can apply the ideal gas law, which states that \( PV = nRT \), where:
- \( P \) is the pressure of the gas,
- \( V \) is the volume,
- \( n \) is the number of moles of gas (which remains constant as particles are added),
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin.
Here’s how we can break it down:
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Initial Conditions:
- The volume is constant, and the initial temperature is 300 K.
- The pressure will fluctuate around some value, \( P_1 \), which we cannot determine without additional information, but we'll consider it a baseline pressure.
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When Heat is Added:
- Adding heat to the container will increase the temperature of the gas. Since the volume is held constant, the pressure will increase as a result of the increased kinetic energy of the particles.
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Lowering Temperature:
- When the temperature is lowered to half its initial value (150 K), according to the ideal gas law, if the volume and the number of particles remain constant, the new pressure \( P_2 \) can be determined by the ratio of the temperatures: \[ P_2 = P_1 \times \frac{T_2}{T_1} = P_1 \times \frac{150 K}{300 K} = 0.5 P_1 \] Thus, the new pressure is about half the original pressure.
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Predicting Pressure with Increased Temperature:
- If the temperature is increased to four times the original temperature (4 × 300 K = 1200 K), then: \[ P_3 = P_1 \times \frac{T_3}{T_1} = P_1 \times \frac{1200 K}{300 K} = 4 P_1 \] Therefore, the pressure would rise to four times the original pressure.
To summarize:
- The pressure fluctuates around \( P_1 \).
- When heat is added, the pressure increases due to increased kinetic energy.
- The new pressure at 150 K is about \( 0.5P_1 \).
- If the temperature is increased to 1200 K, the pressure would rise to \( 4P_1 \).