Chemical Thermodynamics JEE 2025: Coefficient Ratio

What does the Chemical Thermodynamics JEE 2025 question ask?

The chemical thermodynamics JEE 2025 coefficient-ratio question gives a negative answer without solving for molar volume:

- 7.10 m o l d m^{- 3}

It comes from JEE Advanced 2025, Paper 1, Chemistry. “Medium” is the question bank’s difficulty classification, not an official exam rating.

The task is to express the van der Waals equation as a cubic in molar volume, then divide the signed coefficient of the squared term by the signed coefficient of the linear term. The supplied quantities are:

a = 6.0 d m^{6} a t m m o l^{- 2}, b = 0.060 d m^{3} m o l^{- 1}

T = 300 K, P = 300 a t m

R = 0.082 d m^{3} a t m m o l^{- 1} K^{- 1}

This is a numerical-answer question; the supplied record contains no multiple-choice options. The requested ratio has units: $m o l d m^{- 3}$

How do you turn the van der Waals equation into a cubic?

Multiply the molar equation by the square of molar volume, expand, then bring the right-hand side to the left. This follows the official solution and exposes both contributions to the squared-term coefficient. Keep the quantities symbolic until you identify the coefficients.

Start with:

(P + \frac{a}{V_{m}^{2}}) (V_{m} - b) = R T

Multiply both sides by the square of molar volume:

(P + \frac{a}{V_{m}^{2}}) (V_{m} - b) V_{m}^{2} = R T V_{m}^{2}

Expanding the left side gives:

P V_{m}^{3} - P b V_{m}^{2} + a V_{m} - a b = R T V_{m}^{2}

The pressure multiplied by the negative volume correction produces the negative squared term on the left. Moving the right-hand term to the left introduces another negative squared term:

P V_{m}^{3} - P b V_{m}^{2} - R T V_{m}^{2} + a V_{m} - a b = 0

Collect those two terms:

P V_{m}^{3} - (P b + R T) V_{m}^{2} + a V_{m} - a b = 0

Both contributions are negative. One comes from expanding the product; the other comes from moving the right-hand side. They add inside the negative bracket, rather than subtracting from each other.

How do you read the coefficients and calculate the answer?

Divide every term by pressure to make the cubic monic, meaning its leading coefficient is one. Then read the squared and linear coefficients with their signs attached. Their ratio completes the algebra; finding roots would answer a different question.

The normalised cubic is:

V_{m}^{3} - (b + \frac{R T}{P}) V_{m}^{2} + \frac{a}{P} V_{m} - \frac{a b}{P} = 0

Read the signed coefficients separately:

Coefficient of V_{m}^{2} = - (b + \frac{R T}{P})

Coefficient of V_{m} = \frac{a}{P}

The requested ratio is:

\frac{- (b + \frac{R T}{P})}{a / P} = - \frac{P b + R T}{a}

Now substitute the data. Calculate the two numerator contributions separately so that neither is lost:

P b = 300 \times 0.060 = 18 d m^{3} a t m m o l^{- 1}

R T = 0.082 \times 300 = 24.6 d m^{3} a t m m o l^{- 1}

Their units match, so add them directly:

P b + R T = 18 + 24.6 = 42.6 d m^{3} a t m m o l^{- 1}

Divide this sum by the attraction constant. Keep the minus sign outside the calculation.

Final answer:

Required ratio = - \frac{42.6}{6.0} = - 7.10 m o l d m^{- 3}

How can you check the sign, units and normalisation?

The answer must be negative because pressure, temperature, both van der Waals constants and the gas constant are positive here. The sum inside the bracket is positive, while the overall minus sign remains. A positive result fails this check before you repeat any arithmetic.

The dimensional cancellation is:

\frac{d m^{3} a t m m o l^{- 1}}{d m^{6} a t m m o l^{- 2}} = m o l d m^{- 3}

These are coefficient-ratio units, not a calculated gas density, despite matching molar-concentration units. Dividing the whole polynomial by pressure scales both coefficients equally, so the unnormalised coefficients give the same ratio:

\frac{- (P b + R T)}{a} = \frac{- (P b + R T) / P}{a / P}

The question bank’s 120-second expected solve time is its target, not measured student performance. Use expansion, coefficient extraction and substitution, not root-finding.

Why does dropping a coefficient’s sign give a positive answer?

Replacing the signed squared-term coefficient with its magnitude produces a positive result. This is an illustrative wrong numerical result, not an answer option: no options are supplied for this question.

The incorrect extraction is:

Incorrect coefficient of V_{m}^{2} = + (b + \frac{R T}{P})

It should instead be:

Correct coefficient of V_{m}^{2} = - (b + \frac{R T}{P})

The incorrect choice leads to:

+ \frac{P b + R T}{a} = + \frac{42.6}{6.0} = + 7.10 m o l d m^{- 3}

A coefficient includes its algebraic sign. The question asks for a ratio of coefficients, not their magnitudes.

Repair the method in three steps:

Write the polynomial with zero on the right.
Bracket each coefficient together with its sign.
Form the requested ratio only after those signed brackets are written.

Can you solve three related questions without finding molar volume?

All three checks below use the collected cubic and require no roots. These are original practice questions based on this PYQ, not additional verified past-paper questions. Try each before reading its worked check.

Using the original data, what is the numerical monic cubic?

For Question 1, substitute the data into the monic coefficients, retaining the negative squared and constant terms. Calculate the three coefficient quantities:

b + \frac{R T}{P} = 0.060 + \frac{24.6}{300} = 0.142 d m^{3} m o l^{- 1}

\frac{a}{P} = \frac{6.0}{300} = 0.020 d m^{6} m o l^{- 2}

\frac{a b}{P} = \frac{6.0 \times 0.060}{300} = 0.0012 d m^{9} m o l^{- 3}

When molar volume is expressed in cubic decimetres per mole, the numerical cubic is:

V_{m}^{3} - 0.142 V_{m}^{2} + 0.020 V_{m} - 0.0012 = 0

The units above belong to the squared, linear and constant coefficients, respectively. They ensure that every term has the units of molar volume cubed.

What is the ratio of the constant term to the linear coefficient?

For Question 2, the ratio is the negative of the volume correction constant. Use the signed constant term:

\frac{- a b / P}{a / P} = - b = - 0.060 d m^{3} m o l^{- 1}

Pressure and the attraction constant cancel. Temperature does not enter this ratio.

What is the original coefficient ratio if pressure becomes 600 atmospheres?

For Question 3, keep temperature and both van der Waals constants unchanged. Substitute the new pressure into the original ratio:

- \frac{(600 \times 0.060) + (0.082 \times 300)}{6.0} = - \frac{60.6}{6.0} = - 10.10 m o l d m^{- 3}

Increasing pressure at fixed positive constants and temperature makes this signed ratio more negative, because pressure increases the sum under the minus sign. Before redoing the arithmetic, use that direction as your check; no cubic roots are required.

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Frequently asked questions

How do you convert the van der Waals equation into a cubic in molar volume?

Start with (P + a/Vₘ²)(Vₘ − b) = RT and multiply both sides by Vₘ². Expand and collect all terms on the left to obtain PVₘ³ − (Pb + RT)Vₘ² + aVₘ − ab = 0.

What is the answer to the JEE Advanced 2025 thermodynamics coefficient-ratio question?

The ratio of the signed squared-term coefficient to the signed linear coefficient is −(Pb + RT)/a. Substituting the supplied data gives −(18 + 24.6)/6.0 = −7.10 mol dm⁻³.

Why is the coefficient ratio negative rather than positive?

The squared-term coefficient is −(Pb + RT), while the linear coefficient is +a. All the supplied quantities are positive, so their requested ratio must be negative. Using coefficient magnitudes instead of signed coefficients incorrectly gives +7.10 mol dm⁻³.

Do I need to solve the cubic or make it monic to find the coefficient ratio?

Neither step is necessary: read the signed coefficients directly from PVₘ³ − (Pb + RT)Vₘ² + aVₘ − ab = 0. Dividing the entire equation by P makes it monic but scales both coefficients equally, leaving their ratio unchanged. Finding cubic roots would calculate molar volume, which this question does not ask for.