Question

The temperature of a 0.750-l gas sample at 25 °c and 2.00 atm is changed to 250 °c. What is the final pressure of the system, at constant volume?

Answer 1

Final Pressure of the system is 3.51 atm.

Step-by-step explanation:

Initial temperature, T1 = 25 oC = 298.15 K

Initial Pressure, P1 = 2 atm

Final temperature, T2 = 250 oC = 523.15 K

Using the relation;

P1/T1 = P2/T2

P2 = P1T2/T1 = 2*523.15/298.15

P2 = 3.51 atm

Answer 2

The final pressure of the gas sample can be determined using the combined gas law equation: P1 * V1 / T1 = P2 * V2 / T2. Plugging in the given values, the final pressure is calculated to be 9.78 atm.

Step-by-step explanation:

The final pressure of the gas sample can be determined using the combined gas law equation: P1 * V1 / T1 = P2 * V2 / T2. In this case, the initial pressure (P1) is 2.00 atm, the initial volume (V1) is 0.750 L, the initial temperature (T1) is 25 °C + 273.15 K, the final temperature (T2) is 250 °C + 273.15 K, and the final volume (V2) is the same as the initial volume. Plugging in these values into the equation, we can solve for the final pressure (P2).

Using the equation:

P1 * V1 / T1 = P2 * V2 / T2

2.00 atm * 0.750 L / (25 °C + 273.15 K) = P2 * 0.750 L / (250 °C + 273.15 K)

Solving for P2:

P2 = (2.00 atm * 0.750 L / (25 °C + 273.15 K)) * ((250 °C + 273.15 K) / 0.750 L) = 9.78 atm