To analyze the thermodynamic process involving one mole of a monatomic gas transitioning through four distinct states, we can utilize the ideal gas law and the principles of thermodynamics. Let's break down each step and calculate the required parameters: pressure (P), heat flow (q), work done (w), change in internal energy (ΔU), and change in enthalpy (ΔH).
Understanding the Ideal Gas Law
The ideal gas law is given by the equation:
PV = nRT
Where:
- P = pressure in atm
- V = volume in liters
- n = number of moles (1 mole in this case)
- R = ideal gas constant (0.0821 L·atm/(K·mol))
- T = temperature in Kelvin
Step-by-Step Analysis
Step 1: From State A to State B (Constant T)
In this step, the temperature remains constant at 300 K while the volume changes from 10 L to 50 L. Using the ideal gas law:
At State A:
P_a = (nRT) / V = (1 mol * 0.0821 L·atm/(K·mol) * 300 K) / 10 L = 2.463 atm
At State B:
P_b = (nRT) / V = (1 mol * 0.0821 L·atm/(K·mol) * 300 K) / 50 L = 0.493 atm
Step 2: From State B to State C (Constant V)
Here, the volume remains constant at 50 L while the temperature increases to 600 K. The pressure at State C can be calculated as follows:
P_c = (nRT) / V = (1 mol * 0.0821 L·atm/(K·mol) * 600 K) / 50 L = 0.9852 atm
Step 3: From State C to State D (Constant T)
In this step, the temperature remains constant at 600 K while the volume decreases from 50 L to 10 L:
P_d = (nRT) / V = (1 mol * 0.0821 L·atm/(K·mol) * 600 K) / 10 L = 4.911 atm
Step 4: From State D to State A (Constant V)
The volume remains constant at 10 L while the temperature returns to 300 K, which means the pressure will revert to:
P_a = 2.463 atm
Calculating Thermodynamic Quantities
Heat Flow (q)
For an ideal gas, the heat flow can be calculated using:
q = nC_vΔT
Where:
- C_v = (3/2)R for a monatomic gas
- ΔT = change in temperature
For Step 1 (A to B): No heat flow since T is constant.
For Step 2 (B to C):
q = 1 * (3/2)(0.0821)(600 - 300) = 12.315 J
For Step 3 (C to D): No heat flow since T is constant.
For Step 4 (D to A):
q = 1 * (3/2)(0.0821)(300 - 600) = -12.315 J
Work Done (w)
Work done on the system can be calculated using:
w = -PΔV
For Step 1 (A to B):
w = -P_a(V_b - V_a) = -2.463(50 - 10) = -98.52 J
For Step 2 (B to C): No work done since V is constant.
For Step 3 (C to D):
w = -P_c(V_d - V_c) = -0.9852(10 - 50) = 39.408 J
For Step 4 (D to A): No work done since V is constant.
Change in Internal Energy (ΔU)
For an ideal gas, the change in internal energy is given by:
ΔU = nC_vΔT
For Step 1: ΔU = 0 (T is constant).
For Step 2:
ΔU = 1 * (3/2)(0.0821)(600 - 300) = 12.315 J
For Step 3: ΔU = 0 (T is constant).
For Step 4:
ΔU = 1 * (3/2)(0.0821)(300 - 600) = -12.315 J
Change in Enthalpy (ΔH)
The change in enthalpy can be calculated as:
ΔH = nC_pΔT
Where:
- C_p = (5/2)R for a monatomic gas
For Step 1: ΔH = 0 (T is constant).
For Step 2:
ΔH = 1 * (5/2)(0.0821)(600 - 300) = 20.525 J
For Step 3: ΔH = 0 (T is constant).
For Step 4:
ΔH = 1 * (5/2)(0.0821)(300 - 600) = -20.525 J
