One mole of nirtogen at 273K occupying 10 litres (state 1) is to be expanded to a volume of 20 litres at 373K (state 2). An infinite number of paths from state 1 to state 2 are possible. Consider the following important paths:
(a) The gas is expanded isothermally and reversibly at 273K to a volume of 20 litres and then heated to 373K.
(b) The gas at a volume of 10 litres is heated from 273K to 373K and then expanded isothermally to 20 litres.
Compare the numerical values for the change in internal energy, heat absorbed from the surroundings and work done on the gas for the two routes, assuming that nitrogen behaves as an ideal gas and C(v)=(19.0 + 0.005 T/K)J K^-1 mol^-1.
To compare the numerical values for the change in internal energy, heat absorbed from the surroundings, and work done on the gas for the two routes, we need to calculate these values for each path and then compare them.
Let's start with path (a):
1. Calculate the change in internal energy (ΔU):
Since the process is isothermal, the change in internal energy is zero (ΔU = 0). This is because the internal energy of an ideal gas depends only on temperature, and in an isothermal process, the temperature remains constant.
2. Calculate the heat absorbed from the surroundings (Q):
The heat absorbed during an isothermal expansion of an ideal gas can be calculated using the formula Q = nRT ln(V2/V1), where n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, and V1 and V2 are the initial and final volumes, respectively.
Q = (1 mol)(8.314 J/(mol·K))(273 K) ln(20/10)
= 606.48 J
3. Calculate the work done on the gas (W):
The work done during an isothermal expansion of an ideal gas can be calculated using the formula W = -nRT ln(V2/V1), where the negative sign indicates work done by the gas.
W = -(1 mol)(8.314 J/(mol·K))(273 K) ln(20/10)
= -606.48 J
Now let's move on to path (b):
1. Calculate the change in internal energy (ΔU):
We can use the equation ΔU = nCvΔT, where n is the number of moles, Cv is the molar heat capacity at constant volume, and ΔT is the change in temperature.
ΔU = (1 mol)(19.0 + 0.005(373 K - 273 K)) J K^-1 mol^-1
= (1 mol)(19.0 + 0.005(100 K)) J K^-1 mol^-1
= (1 mol)(19.0 + 0.5) J K^-1 mol^-1
= 19.5 J
2. Calculate the heat absorbed from the surroundings (Q):
Since path (b) involves heating the gas from 273K to 373K at constant volume, the heat absorbed can be calculated using the equation Q = ΔU.
Q = ΔU = 19.5 J
3. Calculate the work done on the gas (W):
During an isothermal process, the work done on the gas can be calculated using the formula W = -nRT ln(V2/V1), similar to path (a).
W = -(1 mol)(8.314 J/(mol·K))(273 K) ln(20/10)
= -606.48 J
Comparing the numerical values for both paths:
For path (a):
- Change in internal energy (ΔU) = 0 J
- Heat absorbed from the surroundings (Q) = 606.48 J
- Work done on the gas (W) = -606.48 J
For path (b):
- Change in internal energy (ΔU) = 19.5 J
- Heat absorbed from the surroundings (Q) = 19.5 J
- Work done on the gas (W) = -606.48 J
As we can see, the values for ΔU and Q are lower for path (b) compared to path (a). However, the work done on the gas (W) is the same for both paths, as it depends only on the initial and final volumes.
To compare the numerical values for the change in internal energy, heat absorbed from the surroundings, and work done on the gas for the two routes, we need to calculate these values for each route and then compare them.
(a) Isothermal expansion followed by heating:
In this route, the gas is expanded isothermally and reversibly at 273K to a volume of 20 liters, and then heated to 373K.
1. Change in Internal Energy (ΔU):
Internal energy change during an isothermal process can be calculated using the formula:
ΔU = nRTln(V2/V1)
Given:
n = 1 mole
R = 8.314 J K^-1 mol^-1
T1 = 273K
T2 = 373K
V1 = 10 liters
V2 = 20 liters
Calculating the change in internal energy:
ΔU = (1)(8.314)(273)(ln(20/10))
ΔU = 1.987 J
2. Heat absorbed from the surroundings (Q):
For an isothermal process, Q is equal to the change in internal energy:
Q = ΔU
Q = 1.987 J
3. Work done on the gas (W):
For an isothermal process, work done can be calculated using the formula:
W = -nRTln(V2/V1)
Calculating the work done on the gas:
W = -(1)(8.314)(273)(ln(20/10))
W = -1.987 J
(b) Heating followed by isothermal expansion:
In this route, the gas is heated from 273K to 373K, and then expanded isothermally to 20 liters.
1. Change in Internal Energy (ΔU):
Internal energy change during a heating process can be calculated using the formula:
ΔU = nCvΔT
where Cv is the molar heat capacity at constant volume.
Given:
n = 1 mole
Cv = 19.0 + 0.005T (in J K^-1 mol^-1)
T1 = 273K
T2 = 373K
Calculating the change in internal energy:
ΔU = (1)(19.0 + 0.005(273))(373 - 273)
ΔU = 95.0 J
2. Heat absorbed from the surroundings (Q):
For a heating process, heat absorbed from the surroundings is equal to the change in internal energy:
Q = ΔU
Q = 95.0 J
3. Work done on the gas (W):
For an isothermal process, the work done can be calculated using the formula:
W = -nRTln(V2/V1)
Given:
n = 1 mole
R = 8.314 J K^-1 mol^-1
T = 373K
V1 = 10 liters
V2 = 20 liters
Calculating the work done on the gas:
W = -(1)(8.314)(373)(ln(20/10))
W = -19.9 J
Comparing the values for the two routes:
Route (a):
ΔU = 1.987 J
Q = 1.987 J
W = -1.987 J
Route (b):
ΔU = 95.0 J
Q = 95.0 J
W = -19.9 J
From the calculations, we can see that Route (a) has a smaller change in internal energy and work done on the gas compared to Route (b). The heat absorbed from the surroundings is the same for both routes.