Question

The pressure of a fixed mass of gas is 6.0 x 10^5 Pa. The temperature of the gas is 27°C, and the volume of the gas is 2.5 m³. The temperature of the gas increases to 54°C, and the volume of the gas increases to 5.0 m³. What is the new pressure of the gas?

A. 3.0 x 10^5 Pa
B. 6.0 x 10^5 Pa
C. 12.0 x 10^5 Pa
D. 24.0 x 10^5 Pa

Answer 1

The new pressure of the gas, after the temperature and volume have increased to 54°C and 5.0 m³ respectively, is 6.0 x 10^5 Pa, which remains unchanged due to the proportional increase in both temperature and volume.

Step-by-step explanation:

The student has asked about the new pressure of a fixed mass of gas after its temperature and volume of the gas have been increased. The initial conditions are a pressure of 6.0 x 10^5 Pa, a temperature of 27°C (which is 300.15 K), and a volume of 2.5 m³. After an increase in temperature to 54°C (which is 327.15 K) and volume to 5.0 m³, the new pressure can be calculated using the combined gas law: P1V1/T1 = P2V2/T2. Solving for the new pressure P2 gives:

P2 = P1V1T2 / (T1V2)

P2 = (6.0 x 10^5 Pa * 2.5 m³ * 327.15 K) / (300.15 K * 5.0 m³)

P2 = 6.0 x 10^5 Pa

Thus, the new pressure of the gas is 6.0 x 10^5 Pa, which is option B.