a certain mass of gas occupies 600 cm3 and exerted 1.325*10^5 Nm^-2 pressure. at what temperature would the volume of the gas be halved?
a certain mass of gas occupied 650cm^3 and exerted 1.625×10^5nm^2 pressurf would the volume of the gas halved
To determine the temperature at which the volume of the gas is halved, we can use the combined gas law, which combines Boyle's law, Charles's law, and Gay-Lussac's law. The formula for the combined gas law is:
P1V1 / T1 = P2V2 / T2
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Let's plug in the given values:
P1 = 1.325 * 10^5 Nm^-2 (pressure at the initial volume)
V1 = 600 cm^3 (initial volume)
V2 = V1 / 2 = 600 cm^3 / 2 = 300 cm^3 (final volume)
We need to find T2 (final temperature).
Now, rearrange the equation to solve for T2:
T2 = (P2 * V2 * T1) / (P1 * V1)
Substituting the values:
T2 = (1.325 * 10^5 Nm^-2 * 300 cm^3 * T1) / (1.325 * 10^5 Nm^-2 * 600 cm^3)
Simplifying:
T2 = (300 cm^3 * T1) / (2 * 600 cm^3)
T2 = T1 / 2
Therefore, the final temperature (T2) at which the volume of the gas is halved is equal to half the initial temperature (T1).