An ideal gas has the following initial conditions: Vi = 490 cm3, Pi = 5 atm, and Ti = 100°C. What is its final temperature if the pressure is reduced to 1 atm and the volume expands to 1000 cm3?
______ °C
Pi * Vi / Ti = Pf * Vf / Tf
the T is absolute (Kelvin) temperature
To solve this question, we can use the ideal gas law equation:
PV = nRT,
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the initial temperature from Celsius to Kelvin by adding 273.15:
Ti (Kelvin) = Ti (Celsius) + 273.15 = 100 + 273.15 = 373.15 K.
Next, we can solve for the initial number of moles using the equation:
n = (Pi * Vi) / (R * Ti).
Given:
Pi = 5 atm,
Vi = 490 cm^3,
R = 0.0821 L·atm/(mol·K),
Ti = 373.15 K.
Converting Vi to liters by dividing by 1000:
Vi (liters) = Vi (cm^3) / 1000 = 490 / 1000 = 0.49 L.
Now we substitute the values into the equation to find n:
n = (5 * 0.49) / (0.0821 * 373.15) ≈ 0.062 mol.
Now, let's find the final temperature, Tf.
Rearranging the ideal gas law equation, we get:
Tf = (Pi * Vi) / (n * R).
Given:
Pi = 1 atm,
Vi = 1000 cm^3,
n = 0.062 mol,
R = 0.0821 L·atm/(mol·K).
Converting Vi to liters by dividing by 1000:
Vi (liters) = Vi (cm^3) / 1000 = 1000 / 1000 = 1 L.
Now we substitute the values into the equation to find Tf:
Tf = (1 * 1) / (0.062 * 0.0821) ≈ 19.37 K.
Finally, we convert the temperature back to Celsius by subtracting 273.15:
Tf (Celsius) = Tf (Kelvin) - 273.15 = 19.37 - 273.15 ≈ -253.78°C.
Therefore, the final temperature of the gas is approximately -253.78°C.