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.