the pressure on a gas at-73 Cis doubled, but its volume is held at constant. What will the final temperature be in degrees Cesius?
To calculate the final temperature when the pressure on a gas is doubled, but the volume is held constant, we can use the ideal gas law:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas
In this case, we are keeping the volume constant, so we can simplify the ideal gas law equation to:
P1/T1 = P2/T2
Where:
P1 is the initial pressure of the gas
T1 is the initial temperature of the gas
P2 is the final pressure of the gas
T2 is the final temperature of the gas
Given:
P1 = initial pressure
P2 = 2 * P1 (doubled pressure)
T1 = -73 °C
To find T2, we can rearrange the equation:
P1/T1 = P2/T2
T2 = (P2 * T1) / P1
Plugging in the given values:
T2 = (2 * P1 * T1) / P1
T2 = 2 * T1
Therefore, the final temperature (T2) will be twice the initial temperature (T1).
In this case, since the initial temperature is -73 °C, the final temperature will be (-73 °C) * 2 = -146 °C.
(V1/T1)=(V2/T2)
Don't forget to use T in Kelvin.