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.
To calculate the final temperature of a gas when the pressure and volume change, you can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas (in Pascals)
V is the volume of the gas (in cubic meters)
n is the number of moles of gas
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature of the gas (in Kelvin)
In this case, the gas is monatomic, which means it consists of single atoms, so n (number of moles) is not required for this calculation.
Let's denote the initial temperature as T1, the initial pressure as P1, the initial volume as V1, the final temperature as T2, the final pressure as P2, and the final volume as V2.
We are given the initial temperature (T1 = 20 °C = 293.15 K) and the following changes:
- The pressure is doubled (P2 = 2 * P1)
- The volume is reduced to one-fourth (V2 = (1/4) * V1)
First, let's find the initial pressure and volume in SI units (Pascals and cubic meters, respectively). Since 1 atm = 101325 Pa, and 1 L = 0.001 m^3, we can convert the initial values as follows:
- P1 = 1 atm = 101325 Pa
- V1 = 1 L = 0.001 m^3
Now, let's apply the ideal gas law to the initial and final states:
Initial state: P1 * V1 = T1 * R
Final state: P2 * V2 = T2 * R
We can rearrange the equations to solve for T2:
T2 = (P2 * V2) / R
Substituting the given values:
T2 = (2 * P1 * (1/4) * V1) / R
Now, we can substitute the values of P1, V1, and R:
T2 = (2 * 101325 * 0.25 * 0.001) / 8.314
Simplifying the expression:
T2 = 61.02 K
Therefore, the final temperature of the gas is approximately 61.02 Kelvin.