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If a 3.00-liter tank contains air at 3.00 atm and 20.0°C, and the tank is sealed and cooled until the pressure is 1.00 atm, what will be the final volume of the gas when the temperature is 97.7 Kelvin?

A) 0.69 L
B) 2.08 L
C) 1.15 L
D) 3.45 L

Answer :

Final answer:

By applying the Combined Gas Law, we can find that the final volume of the declined pressure gas would be approximately 0.998 L, considering the variations of pressure and temperature.

Explanation:

To solve this problem, we need to use the Ideal Gas Law and the Combined Gas Law. The Ideal Gas Law states that the pressure of a gas times its volume is equal to the number of moles of the gas times the gas constant times the temperature of the gas. The Combined Gas Law allows us to compare the initial and final states of a gas sample as it undergoes changes in pressure, volume, and temperature. The equation is (P1 x V1)/ T1 = (P2 x V2)/ T2 where P represents pressure, V represents volume, and T represents temperature.

Given P1 = 3 atm, V1 = 3 L, T1 = 20°C (which is 293.15 K when converted to Kelvin), P2 = 1 atm, and T2 = 97.7 K. We rearrange the formula to find V2 which is (P1 x V1 x T2) / (P2 x T1). Plugging these into the formula gives us (3 atm x 3 L x 97.7 K) / (1 atm x 293.15 K) = 0.998 L, which isn't an option provided. There may be a mistake in the provided options or in the initial problem setup.

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