A 3 mg sample of an unknown ideal gas is subjected to an adiabatic and
mechanically reversible expansion from an initial pressure of 2 atmospheres to a final pressure
of 300 torr. Given that gamma
for the gas is 1.532 and that the temperature drop achieved in the
expansion is 231 K, calculate both the initial and final temperatures. Go on to determine
delta H and the molecular weight of the gaseous species if the energy lost in the expansion is
0.126 J. empirical relation
To calculate the initial and final temperatures, we can use the adiabatic expansion equation:
(P1 / T1)^gamma = (P2 / T2)^gamma
Where P1 and P2 are the initial and final pressures respectively, T1 and T2 are the initial and final temperatures respectively, and gamma is the adiabatic exponent.
We are given:
P1 = 2 atm
P2 = 300 torr
gamma = 1.532
Temperature drop = 231 K
First, we need to convert the final pressure from torr to atm:
1 atm = 760 torr
P2 = 300 torr / 760 torr/atm = 0.3947 atm
Now we can rearrange the equation and solve for T2:
(T2 / T1) = (P2 / P1)^(1 / gamma)
(T2 / T1) = (0.3947 atm / 2 atm)^(1 / 1.532)
(T2 / T1) = 0.4125
Next, we can calculate T2 by multiplying T1 by the ratio T2 / T1:
T2 = 0.4125 * T1
We also know that the temperature dropped by 231 K, so we have:
T2 = T1 - 231
Now we can substitute the value of T2 from the second equation into the first equation:
0.4125 * T1 = T1 - 231
Simplifying the equation, we get:
-0.5875 * T1 = -231
Dividing both sides by -0.5875:
T1 = 231 / 0.5875
T1 ≈ 393.19 K
Now that we have the initial temperature (T1), we can calculate the final temperature (T2):
T2 = 0.4125 * T1
T2 ≈ 393.19 K * 0.4125
T2 ≈ 162.02 K
Next, we can calculate delta H (change in enthalpy) using the equation:
delta H = C_p * delta T
Where Cp is the heat capacity at constant pressure and delta T is the temperature change.
Since the process is adiabatic, there is no transfer of heat (Q = 0). Therefore, delta H = 0.
Finally, to determine the molecular weight (M) of the gas, we can use the empirical relation between M and gamma:
M = (gamma - 1) / (gamma + 1)
Substituting the given value of gamma:
M = (1.532 - 1) / (1.532 + 1)
M ≈ 0.532 / 2.532
M ≈ 0.210