To tackle this problem, we need to analyze the behavior of the given diatomic gas under the specified conditions. The gas follows a modified equation of state, which affects its thermodynamic properties. Let's break down the information and derive the necessary values step by step.
Understanding the Equation of State
The equation of state for the gas is given as:
PV = R(T + aT²)
Here, a = 0.001 K-1. This equation indicates that the gas behaves similarly to an ideal gas but has an additional term that accounts for the temperature dependence. This will influence the calculations for work done and heat absorbed.
Work Done on the Gas
When the temperature of the gas is raised at constant pressure, the work done on the gas can be expressed as:
W = P(V₂ - V₁)
For an ideal gas, the work done can be calculated using the ideal gas law, while for our gas, we need to consider the modified equation. The problem states that the work done is 70% higher than that of an ideal gas. If we denote the work done on the ideal gas as Wideal, then:
W = 1.7 * Wideal
Temperature Change and Internal Energy
Next, we need to determine the final temperature T₂ when the initial temperature T₁ = 300 K. The problem suggests that the work done is significantly higher, which implies a larger change in volume or pressure. We can use the relationship between internal energy change and temperature for an ideal gas:
ΔU = nCvΔT
For a diatomic gas, the molar heat capacity at constant volume Cv = (5/2)R. Therefore, the change in internal energy can be expressed as:
ΔU = n(5/2)R(T₂ - T₁)
Calculating Heat Absorbed
The heat absorbed during the process at constant pressure can be calculated using:
Q = nCp
For a diatomic gas, the molar heat capacity at constant pressure is Cp = (7/2)R. Thus:
Q = n(7/2)R(T₂ - T₁)
Finding the Correct Statements
Now, let's analyze the options provided:
- a. T₂ = 400 K, internal energy increases by 250R per mole
- b. T₂ = 400 K, internal energy increases by 350R per mole
- c. Total heat absorbed in process = 450R per mole
- d. Total heat absorbed = 520R per mole
Evaluating Each Option
Assuming T₂ = 400 K:
- ΔT = T₂ - T₁ = 400 K - 300 K = 100 K
- ΔU = n(5/2)R(100) = 250R per mole (which matches option a)
- Q = n(7/2)R(100) = 350R per mole (not matching any option)
Since the work done is 70% higher than the ideal gas, we can conclude that the internal energy change is indeed 250R per mole, which supports option a. However, the heat absorbed does not match any of the options when calculated with T₂ = 400 K.
Final Thoughts
After evaluating the options based on the calculations, it appears that option a is correct regarding the internal energy increase, but the heat absorbed does not match the expected values. Therefore, the most accurate statement based on the calculations is:
Option a: T₂ = 400 K, internal energy increases by 250R per mole.
