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A gas currently occupies 12.2 liters of space at STP. When the gas was cooled, the volume was reduced to 10.1 liters and the pressure was lowered to 0.78 atm. What was the new temperature in Kelvin that caused this change?

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User DxW
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The new temperature in Kelvin that caused the change is 260 K.

Step-by-step explanation:

STP (Standard Temperature and Pressure) is defined as 0°C (273.15 K) and 1 atm pressure. Using the combined gas law formula, we can find the new temperature of the gas when it is cooled and the pressure is lowered.

The combined gas law is:

(P1V1)/T1 = (P2V2)/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature.

At STP, the gas has a pressure of 1 atm, volume of 12.2 L, and temperature of 273.15 K.

When the gas is cooled and the pressure is lowered to 0.78 atm, the volume becomes 10.1 L. We can plug these values into the combined gas law formula to solve for the new temperature:

(1 atm x 12.2 L) / 273.15 K = (0.78 atm x 10.1 L) / T2

Solving for T2, we get:

T2 = (0.78 atm x 10.1 L x 273.15 K) / (1 atm x 12.2 L) = 260 K

Therefore, the new temperature of the gas is 260 K.

User Fluffysheap
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