Chemical Thermodynamics JEE 2023: Compressibility Factor

What is the Chemical Thermodynamics JEE 2023 question and its answer?

The answer to this Chemical Thermodynamics JEE 2023 question is 100: calculate the real-gas pressure first, then use that pressure for the ideal-gas comparison. This is from JEE Advanced 2023, Paper 1, Chemistry, and it is a numerical-answer question, not an MCQ.

At 800 K, the gas has compressibility factor 0.5 and occupies 0.4 cubic decimetres per mole. Its pressure, in atmospheres, is the unknown below:

Z = 0.5, V_{m} = 0.4 d m^{3} m o l^{- 1}, P = x a t m

Find its ideal molar volume at the same temperature and pressure, then calculate the requested ratio:

V_{m, i d e a l} = y d m^{3} m o l^{- 1}, required value = \frac{x}{y}

Use the supplied constant:

R = 8 \times 10^{- 2} L a t m K^{- 1} m o l^{- 1}

The question bank classifies this as medium, with an expected solving time of 90 seconds. That is the question bank’s practice estimate, not an official time limit.

Which equation and units should I use first?

Z = \frac{P V_{m}}{R T}

Use this molar-volume definition for the supplied real gas. Molar volume already means volume per mole, so no additional factor for the number of moles belongs in this equation.

The unit conversion is exact:

1 d m^{3} = 1 L

0.4 d m^{3} m o l^{- 1} = 0.4 L m o l^{- 1}

Keep litres and atmospheres throughout. Converting to cubic metres while retaining the supplied gas constant would mix incompatible units.

R = 0.08 L a t m K^{- 1} m o l^{- 1}

R T = 0.08 \times 800 = 64 L a t m m o l^{- 1}

For the pressure calculation, retain the given compressibility factor. Only the later ideal-gas comparison uses unity.

Z_{r e a l} = 0.5, Z_{i d e a l} = 1

How do I calculate the real-gas pressure?

The real-gas pressure is 80 atm, obtained by retaining the supplied compressibility factor. Calculate pressure first because the ideal comparison must use that same value.

Substitute the real-gas values into the definition:

0.5 = \frac{x \times 0.4}{0.08 \times 800}

Simplify the denominator:

0.5 = \frac{0.4 x}{64}

Multiply both sides by 64: $0.5 \times 64 = 0.4 x$ $32 = 0.4 x$

Divide by 0.4 to obtain the pressure:

x = \frac{32}{0.4} = 80

$P = 80 a t m$ Check this pressure against the original compressibility factor:

\frac{80 \times 0.4}{64} = \frac{32}{64} = 0.5

The supplied real-gas condition is satisfied. Carry 80 atm unchanged into the ideal-gas calculation; the question changes the behaviour assumed, not the pressure or temperature.

How do I find the ideal molar volume and the final ratio?

The ideal molar volume is 0.8 cubic decimetres per mole, giving the required numerical entry 100. Set the compressibility factor to unity while keeping the pressure at 80 atm and temperature at 800 K.

Z_{i d e a l} = 1, P = 80 a t m, T = 800 K

Following the official calculation:

1 = \frac{80 \times y}{0.08 \times 800} = \frac{80 y}{64}

Multiply through, then divide by 80: $80 y = 64$

y = \frac{64}{80} = 0.8

V_{m, i d e a l} = 0.8 L m o l^{- 1} = 0.8 d m^{3} m o l^{- 1}

The requested ratio is:

\frac{x}{y} = \frac{80}{0.8} = 100

Numerical entry: 100. The two unknowns are numerical values in the question’s specified units; pressure divided by molar volume is not dimensionless.

Now check the direction of the volume difference. At equal pressure and temperature:

V_{m, i d e a l} = \frac{R T}{P}

Z = \frac{P V_{m, r e a l}}{R T} = \frac{V_{m, r e a l}}{V_{m, i d e a l}}

Z = \frac{0.4}{0.8} = 0.5

The ideal molar volume must therefore be twice the real molar volume, not half. This compares two descriptions under the same conditions; it does not specify an expansion process or claim that the gas physically changes its behaviour.

Why does an incorrect method give 400 instead of 100?

The result 400 comes from treating the supplied real-gas volume as ideal during the pressure calculation. This numerical-answer question has no supplied options: 400 is an illustrative wrong result, not an official distractor.

The incorrect starting equation for this real gas is: $P V_{m} = R T$

Applied directly to the supplied real-gas volume, it gives:

x_{w r o n g} = \frac{64}{0.4} = 160

Continuing that same incorrect chain:

y_{w r o n g} = \frac{64}{160} = 0.4

\frac{x_{w r o n g}}{y_{w r o n g}} = \frac{160}{0.4} = 400

The arithmetic is consistent, but the starting model is wrong. It silently replaces the given compressibility factor of 0.5 with unity.

Substituting this pressure into the original condition exposes the error:

\frac{160 \times 0.4}{64} = 1 \neq 0.5

Repair the first equation, not the final division. Use the supplied compressibility factor to determine the real-gas pressure first, then switch to unity for the ideal comparison.

What two practice questions can I use to check this method?

Use one question with temperature and molar volume fixed, and another with temperature and pressure fixed. These are original practice questions based on this PYQ, not additional JEE 2023 questions.

What pressure would an ideal gas need to occupy 0.4 litres per mole at 800 K?

It would need 160 atm, using the same gas constant. Here, unlike the original question, ideal behaviour is explicitly required while temperature and molar volume are fixed.

P_{i d e a l} = \frac{R T}{V_{m}} = \frac{64}{0.4} = 160 a t m

Compare this with the original real-gas pressure of 80 atm:

\frac{P_{r e a l}}{P_{i d e a l}} = \frac{80}{160} = 0.5

So 160 atm is correct for this new question. It is incorrect for the original unknown because that gas has compressibility factor 0.5, not unity.

How much smaller is the real volume than the ideal volume at 800 K and 80 atm?

The real volume is 50% smaller, and the recovered compressibility factor is 0.5. Use the given ideal and real molar volumes of 0.8 and 0.4 litres per mole, respectively.

Z = \frac{0.4}{0.8} = 0.5

Percentage reduction = \frac{0.8 - 0.4}{0.8} \times 100 = 50 %

The reference volume is the ideal volume, so it belongs in the percentage denominator. Before using a ratio, write down what stays fixed: pressure ratios require equal temperature and molar volume; volume ratios require equal temperature and pressure.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What is the answer to the JEE Advanced 2023 compressibility factor question?

The numerical answer is 100. Using Z = 0.5 gives the real-gas pressure x = 80 atm, and the ideal molar volume at the same temperature and pressure is y = 0.8 dm³/mol. The requested numerical ratio is x/y = 80/0.8 = 100.

How do I calculate pressure from compressibility factor and molar volume?

Use Z = PVm/RT, rearranged as P = ZRT/Vm. For Z = 0.5, R = 0.08 L atm K⁻¹ mol⁻¹, T = 800 K and Vm = 0.4 L/mol, the pressure is 80 atm. Molar volume is already volume per mole, so no extra factor for the number of moles is needed.

Why is the ideal molar volume twice the real molar volume when Z is 0.5?

At the same temperature and pressure, Z equals the real molar volume divided by the ideal molar volume. With Z = 0.5 and real molar volume 0.4 L/mol, the ideal molar volume is 0.4/0.5 = 0.8 L/mol. The comparison must keep both temperature and pressure unchanged.

Why do I get 400 instead of 100 in this JEE 2023 gas question?

You get 400 if you use PVm = RT for the supplied real gas, ignoring its compressibility factor of 0.5. That incorrect starting equation gives pressure 160 atm and ideal molar volume 0.4 L/mol. Use P = ZRT/Vm first to obtain 80 atm, then use ideal behaviour only for the molar-volume comparison.