To find the change in temperature (ΔT) of the ideal gas, we follow a step-by-step thermodynamic and electrochemical derivation:
Step 1: Standard EMF of the Fuel Cell
The standard reduction potentials for the two half-cells are given as:
Cathode (Reduction): O2(g)+4H+(aq)+4e−→2H2O(l),Ecathode0=1.23 V
Anode (Oxidation): H2(g)→2H+(aq)+2e−,Eanode0=0.00 V
The standard cell potential (Ecell0) is:
Ecell0=Ecathode0−Eanode0=1.23 V−0.00 V=1.23 V
Step 2: Maximum Electrical Work Available
For the cell reaction:
H2(g)+21O2(g)→H2O(l)
The number of moles of electrons transferred per mole of H2(g) consumed is n=2.
The maximum theoretical work (Wmax) derived from consuming nH2=1.0×10−3 mol of H2(g) is:
Wmax=−ΔG0=nH2⋅n⋅F⋅Ecell0
Substituting the given values:
Wmax=(1.0×10−3 mol)×2×96500 C mol−1×1.23 V
Wmax=193 C×1.23 V=237.39 J
Step 3: Actual Work Derived from the Fuel Cell
Since the efficiency (η) of the fuel cell is 70%:
Wactual=η×Wmax=0.70×237.39 J=166.173 J
Step 4: Temperature Change of the Ideal Gas
The work derived from the cell is used to compress 1.00 mol of a monoatomic ideal gas in a thermally insulated container.
Since the container is thermally insulated, the process is adiabatic (q=0). By the First Law of Thermodynamics:
ΔU=q+Won gas=Wactual
For 1.00 mol of a monoatomic ideal gas:
ΔU=ngas⋅Cv,m⋅ΔT
For a monoatomic ideal gas, Cv,m=23R:
Cv,m=23×8.314 J mol−1 K−1=12.471 J mol−1 K−1
Equating ΔU to Wactual:
1.00 mol×12.471 J mol−1 K−1×ΔT=166.173 J
ΔT=12.471166.173≈13.32 K
Thus, the change in temperature of the ideal gas is 13.32 K (or within the range 13.32−13.33 K).