Thermodynamics JEE 2025: Water Electrolysis Work Solved

What is the answer to the Thermodynamics JEE 2025 water-electrolysis question?

The accepted answer is 29.88 kJ, the magnitude of expansion work. This thermodynamics JEE 2025 question belongs to JEE Advanced 2025, Paper 1, Chemistry. It is a numerical-answer question, not a JEE Main question.

The task is to find the magnitude of expansion work in kJ when 144 g of water undergoes complete electrolysis at 300 K under constant pressure, producing ideal gases. The question bank rates it medium. Use the supplied gas constant and atomic masses:

R = 8.3 J K^{- 1} m o l^{- 1}, A_{H} = 1, A_{O} = 16

How does 144 g of water produce 12 mol of gas?

The water contains 8 mol of liquid reactant, but electrolysis produces 12 mol of gaseous products. Each two moles of water give three moles of gases, so convert through the balanced reaction before applying the ideal-gas equation.

First, calculate the molar mass and the amount of water:

M (H_{2} O) = 2 (1) + 16 = 18 g m o l^{- 1}

n (H_{2} O) = \frac{144}{18} = 8 m o l

Complete electrolysis follows:

2 H_{2} O (l) \to 2 H_{2} (g) + O_{2} (g)

Therefore, the total gas amount is:

n_{g a s} = 8 \times \frac{3}{2} = 12 m o l

Check the two products separately:

n (H_{2}) = 8 m o l, n (O_{2}) = 4 m o l

n_{g a s} = 8 + 4 = 12 m o l

Both hydrogen and oxygen contribute to the final gas volume. Counting only hydrogen misses the oxygen contribution, even though the hydrogen amount equals the initial water amount.

How do you calculate expansion work without knowing the pressure?

The pressure cancels when the final gas volume is substituted into the work equation. Following the official solution, neglect the initial liquid-water volume compared with the final gas volume. This is an approximation about relative volumes, not a claim that liquid water occupies no volume.

For the final ideal gases: $P V = n R T$

V_{f i n a l} = \frac{n R T}{P} = \frac{12 \times 8.3 \times 300}{P}

The volume change is:

Δ V = V_{f i n a l} - V_{i n i t i a l} \approx \frac{12 \times 8.3 \times 300}{P} - 0

Use the chemistry convention for pressure–volume work: $w = - P_{e x t} Δ V$

The official solution identifies the constant external pressure with the pressure used in the final gas equation. Constant external pressure alone does not establish this equality in every process: $P_{e x t} = P$

Substituting gives:

w = - P [\frac{12 \times 8.3 \times 300}{P} - 0] = - 12 \times 8.3 \times 300 J

No numerical pressure is required because the same pressure multiplies the volume and appears in its denominator. Use this constant-pressure boundary-work route, not a reversible-isothermal logarithmic formula or an electrical-work calculation.

Why is the numerical answer 29.88 rather than −29.88?

The question requires the magnitude, while chemistry's signed work on the system is negative. During expansion, the system does work on the surroundings. The negative sign records that direction of energy transfer.

Complete the arithmetic: $12 \times 8.3 = 99.6$ $99.6 \times 300 = 29, 880$

Hence the signed work on the system is:

w = - 29, 880 J = - 29.88 k J

The official solution interprets the requested expansion work as its magnitude: $| w | = 29.88 k J$

Enter 29.88 in the numerical-answer field. This is not a rule that all work answers must be positive. If a question asks for signed chemistry work, retain the negative sign for expansion.

Why does using 8 mol give the incorrect result 19.92 kJ?

Using 8 mol directly in the gas equation counts liquid reactant instead of total gaseous products. This is a numerical-answer question, so 19.92 kJ is an illustrative incorrect result, not a supplied option.

The faulty substitution is: $n = 8 m o l$

| w | = 8 \times 8.3 \times 300 = 19, 920 J = 19.92 k J

Ask the diagnostic question: “Which species actually occupy the final gas volume?” Here, both hydrogen and oxygen do. The water calculation gives the starting liquid amount, whereas the ideal-gas equation needs the combined amount of gaseous products.

Repair the method by inserting the balanced-reaction conversion. Label the species beside each mole calculation before substituting into the gas equation:

n_{g a s} = 8 \times \frac{3}{2} = 12 m o l

Can you solve three variations using the same method?

Convert water mass to total gas moles, check whether the boundary moves, then calculate the requested work. These are original practice variations, not additional verified PYQs. Reuse the supplied atomic masses and gas constant. For the first two, retain the official solution's pressure identification and negligible initial liquid-volume approximation.

What is the expansion-work magnitude for complete electrolysis of 36 g of water at 300 K?

The magnitude is 7.47 kJ under the same constant-pressure assumptions. This tests mass-to-gas conversion, so write both mole calculations:

n_{w a t e r} = \frac{36}{18} = 2 m o l

n_{g a s} = 2 \times \frac{3}{2} = 3 m o l

| w | = \frac{3 \times 8.3 \times 300}{1000} = 7.47 k J

What changes if 144 g of water is completely electrolysed at 350 K?

The gas amount remains 12 mol, but the work magnitude increases to 34.86 kJ. This tests temperature dependence at unchanged gas amount under the same assumptions.

n_{g a s} = \frac{144}{18} \times \frac{3}{2} = 12 m o l

| w | = \frac{12 \times 8.3 \times 350}{1000} = 34.86 k J

w = - 34.86 k J

What is the pressure–volume boundary work for electrolysis in a sealed rigid vessel?

The pressure–volume boundary work is zero because the vessel's boundary does not move. This tests the need for an actual boundary-volume change, not merely gas production.

Δ V = 0 \Rightarrow w_{P V} = - P_{e x t} Δ V = 0

This does not mean electrical work is zero. Keep those energy-transfer modes separate.

Next, practise another question from the chapter: Chemical Thermodynamics JEE 2025: Coefficient Ratio.

Frequently asked questions

What is the answer to the JEE Advanced 2025 water-electrolysis question?

The accepted numerical answer is 29.88, the expansion-work magnitude in kJ. Under the chemistry sign convention, the signed work on the system is −29.88 kJ because the system expands. Enter the positive value because this question requires the magnitude.

Why do we use 12 mol instead of 8 mol for water electrolysis?

The 144 g of water contains 8 mol of liquid water, but complete electrolysis produces 8 mol of hydrogen and 4 mol of oxygen. The ideal-gas equation therefore needs 12 mol of gaseous products. Using 8 mol gives the incorrect work magnitude of 19.92 kJ.

How can expansion work be calculated without knowing the pressure?

The official solution neglects the initial liquid-water volume and equates the constant external pressure with the final gas pressure. Substituting V = nRT/P into w = −PΔV then gives w ≈ −nRT, so pressure cancels. Constant external pressure alone does not justify that pressure equality in every process.

Is expansion work zero when water is electrolysed in a rigid vessel?

Pressure–volume boundary work is zero in a sealed rigid vessel because its boundary volume does not change. Thus, ΔV = 0 gives w = −P_extΔV = 0. This does not mean that electrical work is zero.

Thermodynamics JEE 2025: Water Electrolysis Work