Answer:
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To calculate the various quantities for the isothermal expansion of the ideal gas, we can use the equations related to the First Law of Thermodynamics and the Second Law of Thermodynamics.
Given:
Initial pressure (P₁) = 3.00 atm
Final pressure (P₂) = 1.00 atm
External pressure (P_ext) = 1.00 atm
Number of moles (n) = 1.00 mol
Temperature (T) = 37°C (convert to Kelvin: T = 37 + 273.15 = 310.15 K)
a) The heat (q):
Since the process is isothermal (constant temperature), the heat exchanged can be calculated using the equation:
q = nRT ln(P₂/P₁)
where R is the ideal gas constant.
Plugging in the values:
q = (1.00 mol)(0.0821 L·atm/(mol·K))(310.15 K) ln(1.00 atm / 3.00 atm)
Calculating:
q = -12.42 J (rounded to two decimal places)
b) The work (w):
The work done during an isothermal expansion can be calculated using the equation:
w = -nRT ln(V₂/V₁)
where V is the volume of the gas.
Since the process is against a constant external pressure, the work done is given by:
w = -P_ext(V₂ - V₁)
Since the external pressure is constant at 1.00 atm, the work can be calculated as:
w = -1.00 atm (V₂ - V₁)
c) The change in internal energy (ΔU):
For an isothermal process, the change in internal energy is zero:
ΔU = 0
d) The change in enthalpy (ΔH):
Since the process is isothermal, the change in enthalpy is equal to the heat (q):
ΔH = q = -12.42 J
e) The change in entropy of the system (ΔS_sys):
The change in entropy of the system can be calculated using the equation:
ΔS_sys = nR ln(V₂/V₁)
Since it's an isothermal process, the change in entropy can also be calculated as:
ΔS_sys = q/T
Plugging in the values:
ΔS_sys = (-12.42 J) / (310.15 K)
Calculating:
ΔS_sys = -0.040 J/K (rounded to three decimal places)
f) The change in entropy of the surroundings (ΔS_sur):
Since the process is reversible and isothermal, the change in entropy of the surroundings is equal to the negative of the change in entropy of the system:
ΔS_sur = -ΔS_sys = 0.040 J/K (rounded to three decimal places)
g) The total change in entropy (ΔS_total):
The total change in entropy is the sum of the changes in entropy of the system and the surroundings:
ΔS_total = ΔS_sys + ΔS_sur = -0.040 J/K + 0.040 J/K = 0 J/K
Therefore, the answers are:
a) q = -12.42 J
b) w = -1.00 atm (V₂ - V₁)
c) ΔU = 0
d) ΔH = -12.42 J
e) ΔS_sys = -0.040 J/K
f) ΔS_sur = 0.040 J/K
g) ΔS_total = 0 J/K