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A canister containing air has a volume of 8.5 L and a pressure of 1.45 atm when the temperature is 310 K. What is the pressure when the volume is increased to 18.0 L and the temperature is reduced to 280 K?

User Racky
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1 Answer

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Final answer:

To find the final pressure when the volume is increased and the temperature is reduced, we can use the ideal gas law. By solving for the number of moles using the initial conditions and then plugging in the final conditions, we can calculate the new pressure. In this case, the final pressure is approximately 1.223 atm.

Step-by-step explanation:

To solve this problem, we can use the ideal gas law, which is represented by the equation PV = nRT. P represents the pressure, V represents the volume, n represents the number of moles of gas, R is the ideal gas constant, and T represents the temperature in Kelvin.

First, let's find the initial number of moles of gas. We rearrange the ideal gas law equation to solve for n: n = PV/RT.

Using the given values:

Initial volume (V1) = 8.5 L

Initial pressure (P1) = 1.45 atm

Initial temperature (T1) = 310 K

Plugging these values into the equation, we can find n1:

n1 = (1.45 atm * 8.5 L) / (0.0821 atm·L/mol·K * 310 K) ≈ 0.474 mol

Now, let's use the ideal gas law to find the final pressure (P2) when the volume (V2) is increased to 18.0 L and the temperature (T2) is reduced to 280 K.

Using the equation P1V1/T1 = P2V2/T2, we can rearrange to solve for P2:

P2 = (P1 * V1 * T2) / (V2 * T1)

Plugging in the given values:

P1 = 1.45 atm

V1 = 8.5 L

T1 = 310 K

V2 = 18.0 L

T2 = 280 K

We can now calculate the final pressure (P2):

P2 = (1.45 atm * 8.5 L * 280 K) / (18.0 L * 310 K) ≈ 1.223 atm

Therefore, the pressure when the volume is increased to 18.0 L and the temperature is reduced to 280 K is approximately 1.223 atm.

User Magras
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