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?
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To find the final temperature of the ideal 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
First, let's convert the initial volume, Vi, from cm^3 to liters:
Vi = 490 cm^3 * (1 L / 1000 cm^3) = 0.49 L
Next, we need to convert the initial temperature, Ti, from Celsius to Kelvin:
Ti = 100°C + 273.15 = 373.15 K
We can rewrite the ideal gas equation as:
(Pi)(Vi) / Ti = (Pf)(Vf) / Tf
Where:
Pi = initial pressure = 5 atm (given)
Vi = initial volume = 0.49 L (calculated)
Ti = initial temperature = 373.15 K (calculated)
Pf = final pressure = 1 atm (given)
Vf = final volume = 1 L (converted from cm^3)
Tf = final temperature (to be determined)
Now we can rearrange the equation to solve for Tf:
Tf = (Pf)(Vf)(Ti) / (Pi)(Vi)
Tf = (1 atm)(1 L)(373.15 K) / (5 atm)(0.49 L)
Tf = 373.15 K / (5 * 0.49)
Tf = 373.15 K / 2.45
Tf = 152.24 K
Therefore, the final temperature is 152.24°C.