Two moles of neon gas at 25oC and 2.0 atm is expanded to 3 times the original volume while the pressure is reduced to 1.0 atm. Find the end temperature.
A. 447degreesC B. 174 degreesC C. -66 degrees C D. 38degreesC E. 150degreesC
To find the end temperature of the neon gas, 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
R is the ideal gas constant
T is the temperature
In this problem, we have the initial conditions (2 moles of neon gas at 25 degrees Celsius and 2.0 atm pressure) and we need to find the final temperature after the gas is expanded to 3 times the original volume while the pressure is reduced to 1.0 atm.
First, we need to convert the initial temperature from Celsius to Kelvin:
T(initial) in Kelvin = T(initial in Celsius) + 273.15
T(initial) in Kelvin = 25 + 273.15 = 298.15 K
Using the ideal gas law equation for the initial conditions:
P(initial) * V(initial) = n * R * T(initial)
Substituting the values:
(2.0 atm) * V(initial) = (2 moles) * (0.0821 L*atm/(mol*K)) * (298.15 K)
Simplifying further:
V(initial) = (2 * 0.0821 * 298.15) / 2.0
V(initial) = 24.457 L
After the gas is expanded to 3 times the original volume, the final volume is:
V(final) = 3 * V(initial)
V(final) = 3 * 24.457 L
V(final) = 73.371 L
Now we can use the ideal gas law equation again for the final conditions:
P(final) * V(final) = n * R * T(final)
Substituting the known values:
(1.0 atm) * (73.371 L) = (2 moles) * (0.0821 L*atm/(mol*K)) * T(final)
Simplifying further:
73.371 = 0.1642 * T(final)
Dividing both sides by 0.1642:
T(final) ≈ 447.14 K
Finally, we need to convert the temperature from Kelvin back to degrees Celsius:
T(final in Celsius) = T(final in Kelvin) - 273.15
T(final in Celsius) ≈ 447.14 - 273.15 ≈ 173.99 ≈ 174 degrees Celsius
Therefore, the answer is B. 174 degrees Celsius.