Final answer:
Using Boyle's Law and the constant volume of the gas, the final pressure of ethane in the flask after heating and then cooling is equal to the initial atmospheric pressure, which is 1.013 \u00d7 105 Pa or 1 atm.
Step-by-step explanation:
To find the final pressure of ethane in the flask after it is warmed to 550 K and then cooled back to 300 K with the stopcock closed, we can use the combined gas law, which connects the pressure, volume, and temperature of a gas. The combined gas law is represented by the formula: (P1V1/T1) = (P2V2/T2), where P is pressure, V is volume, and T is temperature in Kelvins. However, since the volume and the amount of gas are constant, this simplifies to Boyle's Law (P1/T1 = P2/T2).
To solve the problem, we follow these steps:
- Convert pressure to atmospheres (1 atm = 1.013 \u00d7 105 Pa) if necessary.
- Use the initial conditions as P1 and T1 (300 K, in this case).
- Since the temperature increases to 550 K with the stopcock open, the pressure remains constant at atmospheric pressure, meaning P2 = P1 at this point.
- Now, close the stopcock and bring the gas back to its original temperature of 300 K to determine the final pressure, P3. We use P1/T1 = P3/T3 since the volume is constant.
- Solve for P3 using the calculation P3 = P1 (T3/T1).
Given that the initial pressure, P1, is known, and both initial and final temperatures are given (T1 = 300 K, T3 = 300 K), we plug these values into the equation and solve for the final pressure, P3, which will be equal to the initial pressure, P1, because the temperature ratio T3/T1 is 1.
The final pressure of the ethane in the flask would be the same as the initial pressure, 1.013 \u00d7 105 Pa or 1 atm, since both the initial and the final temperatures are 300 K.