Question

A sample of a gas originally at 29 degree celsius and 1.25 atm in 3.0 L container is allowed to contract until the volume is 2.2L and the temperature is 11 degree celsius. The final pressure of the gas is __________ atm.

Answer 1

By applying the combined gas law, the final pressure of the gas after it contracts to 2.2L and cools to 11 degrees Celsius is calculated to be 1.474 atm.

Step-by-step explanation:

The problem described in the question is a classic application of the combined gas law, which allows us to calculate changes in pressure, volume, and temperature for a gas sample when two sets of conditions are known. The combined gas law is expressed as:

P1V1/T1 = P2V2/T2

For the given problem, we have the initial conditions (P1 = 1.25 atm, V1 = 3.0 L, T1 = 29 + 273.15 = 302.15 K) and the final conditions after the change (V2 = 2.2 L, T2 = 11 + 273.15 = 284.15 K). The final pressure P2 is what we want to solve for.

By rearranging the combined gas law to solve for P2, we get:

P2 = (P1V1T2) / (T1V2)

Plugging in the given values yields:

P2 = (1.25 atm * 3.0 L * 284.15 K) / (302.15 K * 2.2 L) = 1.474 atm

So the final pressure of the gas, after it is allowed to contract and the temperature changes, is 1.474 atm.