JEE Challenger
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Temperature Change of Ideal Gas Compressed by Fuel Cell Work

A hydrogen-oxygen fuel cell operates with an efficiency of 70%70\% under standard conditions (1 bar1\text{ bar} and 298 K298\text{ K}). The overall cell reaction is: H2(g)+12O2(g)→H2O(l)H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)

The work derived from the cell upon consuming 1.0×10−3 mol1.0 \times 10^{-3}\text{ mol} of H2(g)H_2(g) is used to compress 1.00 mol1.00\text{ mol} of a monoatomic ideal gas contained in a thermally insulated vessel. Determine the change in temperature (in K) of the ideal gas.

The standard reduction potentials for the half-cells are given as: O2(g)+4H+(aq)+4e−→2H2O(l),E0=1.23 VO_2(g) + 4H^+(aq) + 4e^- \rightarrow 2 H_2O(l), \quad E^0 = 1.23\text{ V} 2H+(aq)+2e−→H2(g),E0=0.00 V2H^+(aq) + 2e^- \rightarrow H_2(g), \quad E^0 = 0.00\text{ V}

(Use Faraday's constant F=96500 C mol−1F = 96500\text{ C mol}^{-1} and gas constant R=8.314 J mol−1 K−1R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}.)

Official Numerical Answer13.32 to 13.33

Step-by-Step Solution

To find the change in temperature (ΔT\Delta T) 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\text{Cathode (Reduction): } O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2 H_2O(l), \quad E^0_{\text{cathode}} = 1.23\text{ V} Anode (Oxidation): H2(g)→2H+(aq)+2e−,Eanode0=0.00 V\text{Anode (Oxidation): } H_2(g) \rightarrow 2H^+(aq) + 2e^-, \quad E^0_{\text{anode}} = 0.00\text{ V}

The standard cell potential (Ecell0E^0_{\text{cell}}) is: Ecell0=Ecathode0−Eanode0=1.23 V−0.00 V=1.23 VE^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} = 1.23\text{ V} - 0.00\text{ V} = 1.23\text{ V}

Step 2: Maximum Electrical Work Available

For the cell reaction: H2(g)+12O2(g)→H2O(l)H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)

The number of moles of electrons transferred per mole of H2(g)H_2(g) consumed is n=2n = 2.

The maximum theoretical work (WmaxW_{\text{max}}) derived from consuming nH2=1.0×10−3 moln_{H_2} = 1.0 \times 10^{-3}\text{ mol} of H2(g)H_2(g) is: Wmax=−ΔG0=nH2⋅n⋅F⋅Ecell0W_{\text{max}} = -\Delta G^0 = n_{H_2} \cdot n \cdot F \cdot E^0_{\text{cell}}

Substituting the given values: Wmax=(1.0×10−3 mol)×2×96500 C mol−1×1.23 VW_{\text{max}} = (1.0 \times 10^{-3}\text{ mol}) \times 2 \times 96500\text{ C mol}^{-1} \times 1.23\text{ V} Wmax=193 C×1.23 V=237.39 JW_{\text{max}} = 193\text{ C} \times 1.23\text{ V} = 237.39\text{ J}

Step 3: Actual Work Derived from the Fuel Cell

Since the efficiency (η\eta) of the fuel cell is 70%70\%: Wactual=η×Wmax=0.70×237.39 J=166.173 JW_{\text{actual}} = \eta \times W_{\text{max}} = 0.70 \times 237.39\text{ J} = 166.173\text{ J}

Step 4: Temperature Change of the Ideal Gas

The work derived from the cell is used to compress 1.00 mol1.00\text{ mol} of a monoatomic ideal gas in a thermally insulated container.

Since the container is thermally insulated, the process is adiabatic (q=0q = 0). By the First Law of Thermodynamics: ΔU=q+Won gas=Wactual\Delta U = q + W_{\text{on gas}} = W_{\text{actual}}

For 1.00 mol1.00\text{ mol} of a monoatomic ideal gas: ΔU=ngas⋅Cv,m⋅ΔT\Delta U = n_{\text{gas}} \cdot C_{v,m} \cdot \Delta T

For a monoatomic ideal gas, Cv,m=32RC_{v,m} = \frac{3}{2}R: Cv,m=32×8.314 J mol−1 K−1=12.471 J mol−1 K−1C_{v,m} = \frac{3}{2} \times 8.314\text{ J mol}^{-1}\text{ K}^{-1} = 12.471\text{ J mol}^{-1}\text{ K}^{-1}

Equating ΔU\Delta U to WactualW_{\text{actual}}: 1.00 mol×12.471 J mol−1 K−1×ΔT=166.173 J1.00\text{ mol} \times 12.471\text{ J mol}^{-1}\text{ K}^{-1} \times \Delta T = 166.173\text{ J}

ΔT=166.17312.471≈13.32 K\Delta T = \frac{166.173}{12.471} \approx 13.32\text{ K}

Thus, the change in temperature of the ideal gas is 13.32 K13.32\text{ K} (or within the range 13.32−13.33 K13.32 - 13.33\text{ K}).