A sample of a monatomic ideal gas is originally at 20 °C.
Calculate its final temperature of the gas if the pressure is
doubled and volume is reduced to one-fourth its initial value.
PV = kT
(2P)(V/4) = k(T/2)
10
To calculate the final temperature of the gas, we can use the ideal gas law, which states that:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature.
In this case, we can assume that the number of moles of the gas remains constant.
Let's label the initial conditions as follows:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
And the final conditions as:
P2 = final pressure
V2 = final volume
T2 = final temperature
We are given:
P1 = initial pressure = P
V1 = initial volume = V
T1 = initial temperature = 20 °C
We are also given:
P2 = final pressure = 2 * P1 (pressure is doubled)
V2 = final volume = V1 / 4 (volume is reduced to one-fourth)
To find T2, we need to solve for it using the ideal gas law equation.
First, let's convert the initial temperature from °C to Kelvin (K):
T1 = 20 °C + 273.15 = 293.15 K
Now, we can set up the equation:
P1V1 / T1 = P2V2 / T2
Substituting the given values:
P * V / 293.15 = (2 * P * (V / 4)) / T2
Canceling out the P factor:
V / 293.15 = (2 * V / 4) / T2
Simplifying:
V / 293.15 = V / (2 * T2)
Cross multiplying:
V * (2 * T2) = V * 293.15
Dividing by V:
2 * T2 = 293.15
Finally, solving for T2:
T2 = 293.15 / 2 = 146.575 K
Therefore, the final temperature of the gas would be approximately 146.575 K.
To calculate the final temperature of the gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
Since the gas is monatomic, the number of moles (n) will remain constant throughout the process. Therefore, we can rewrite the equation as:
P1V1 / T1 = P2V2 / T2
Given:
T1 = 20 °C = 20 + 273.15 = 293.15 K (Converting from Celsius to Kelvin)
To find T2, we need to determine the new values for P2 and V2.
According to the problem:
P2 = 2P1 (pressure is doubled)
V2 = V1/4 (volume is reduced to one-fourth)
Substituting these values into the equation:
(2P1)(V1 / 4) / T2 = P1V1 / T1
Simplifying:
P1V1 / 2T2 = P1V1 / T1
Canceling out P1V1:
1/2T2 = 1/T1
Cross multiplying:
2T1 = T2
Substituting in the known values:
2(293.15 K) = T2
T2 = 586.30 K
Therefore, the final temperature of the gas, when the pressure is doubled and the volume is reduced to one-fourth, is 586.30 K.