According to Gay-Lussac's law (at constant volume), what happens to pressure when temperature increases?

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Multiple Choice

According to Gay-Lussac's law (at constant volume), what happens to pressure when temperature increases?

Explanation:
Direct proportionality of pressure to temperature at constant volume is what Gay-Lussac's law describes. As temperature rises, gas molecules move faster and collide with container walls more forcefully, so the pressure increases. For an ideal gas, P is proportional to T when volume is fixed, so P2 = P1(T2/T1) with temperatures in Kelvin. Using Kelvin is essential because 0 K corresponds to zero molecular motion and provides a meaningful zero point. Therefore, increasing temperature results in a higher pressure, in direct proportion. The other statements—pressure decreasing, staying the same, or being independent of temperature—do not align with this relationship.

Direct proportionality of pressure to temperature at constant volume is what Gay-Lussac's law describes. As temperature rises, gas molecules move faster and collide with container walls more forcefully, so the pressure increases. For an ideal gas, P is proportional to T when volume is fixed, so P2 = P1(T2/T1) with temperatures in Kelvin. Using Kelvin is essential because 0 K corresponds to zero molecular motion and provides a meaningful zero point. Therefore, increasing temperature results in a higher pressure, in direct proportion. The other statements—pressure decreasing, staying the same, or being independent of temperature—do not align with this relationship.

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