JC1 H2 Chemistry Tutor | Preparing an Ideal-Gas Task with Absolute Pressure and Consistent Units

Write the outcome as a connected calculation: identify a gas sample’s pressure, volume and absolute temperature, select a consistent gas constant and determine the requested amount or property. Then state the assumptions under which the result is meaningful.

Supply a fixed amount of gas in thermal equilibrium at the stated temperature. If the gas reacts, leaks, condenses or changes composition during the measurement, the simple inventory needs further information. Do not silently treat a changing sample as unchanged.

The ideal-gas model treats molecular volume as negligible relative to the container volume and neglects intermolecular attractions in the equation of state. It is a model with conditions of useful approximation, not a claim that real molecules have no size or interactions.

Ask the student what each symbol represents before rearranging: p is absolute pressure, V is the gas volume, n is amount in moles and T is thermodynamic temperature. R carries the unit convention that connects them.

Prepare a quantity table with an original value, converted value and unit for each input. A short table can reveal whether the error occurs before the equation is used and can be removed as the student gains independent control.

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Supply the main pressure as 100 kPa absolute. In SI units this is 100 000 Pa. The word absolute prevents confusion with a pressure measured relative to the surrounding atmosphere.

For a separate gauge example, state gauge pressure 100 kPa and atmospheric pressure 100 kPa. The absolute pressure is 200 kPa. Using only the gauge reading in pV = nRT would omit the atmospheric contribution in that supplied case.

Use T = 300 K in the main dataset. If another question supplies Celsius temperature, convert using the stated relation T/K = t/°C + 273.15, with rounding appropriate to the task. Temperature ratios in the gas equation use kelvin.

A Celsius value of 27 is not interchangeable with approximately 300 K. Substituting 27 into the main equation would represent a very different thermodynamic temperature, not a slightly different unit label.

Ask whether the pressure and temperature are final equilibrium conditions or readings from different stages. The equation must use a consistent state of the sample. A pressure measured before warming cannot be combined casually with the final volume and temperature.

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Give V = 2.50 dm³ for the main gas sample. With R = 8.31 J mol⁻¹ K⁻¹ and pressure in Pa, convert to 2.50 × 10⁻³ m³. The product Pa m³ is equivalent to joules.

The conversion from dm³ to m³ is a factor of 10⁻³, not 10⁻¹. Volume scales with the cube of the length conversion. Likewise, 1 cm³ is 10⁻⁶ m³ and 1000 cm³ is 1 dm³.

An equally consistent convention uses pressure in kPa and volume in dm³ with R = 8.31 kPa dm³ mol⁻¹ K⁻¹. One kPa dm³ equals one joule, so the same numerical R works with that paired unit system.

This does not permit every arbitrary pressure and volume combination with 8.31. If another R is supplied in different units, use its stated convention. The tutor should audit the units rather than insist on one memorised conversion pathway.

Ask the student to compare the two valid products for this example: 100 000 × 0.00250 Pa m³ and 100 × 2.50 kPa dm³. Both represent the same 250 J pressure–volume product.

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Rearrange to n = pV/(RT). Using SI values gives n = (100 000 × 0.00250)/(8.31 × 300) = 250/2493 = 0.10028 mol, reported as 0.100 mol to three significant figures.

Supply a gas mass of 4.40 g belonging to that same dry sample. Its molar mass is M = m/n = 4.40/0.10028 = 43.9 g mol⁻¹ after appropriate rounding. Retain guard digits in n until the final calculation.

The relative molecular mass has the corresponding numerical value 43.9 on the relative scale and has no unit. Distinguish it from molar mass, which is expressed here in g mol⁻¹.

An approximate molar mass near 44 does not uniquely identify a gas. Different molecules can have similar molar masses, and an unrecognised mixture can yield an apparent mean value. Additional chemical or spectroscopic evidence would be needed for identity.

Ask the student to state which measurement each calculation uses. The amount follows from pressure, volume, temperature and the model; the molar mass then combines that amount with the sample mass. This prevents unrelated masses being imported into a correct gas calculation.

The same inputs give an ideal molar volume RT/p = 24.93 dm³ mol⁻¹ at 300 K and 100 kPa. A memorised 24 dm³ mol⁻¹ is not an exact universal gas volume. If an examination supplies a particular molar volume for stated conditions, use it within that question’s specified model.

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The table records the original and SI inputs for the main task. Keep it beside the key so that a wrong answer can be traced to a conversion, a rearrangement or a model assumption rather than marked as an undifferentiated calculation error.

The pressure–volume product is 250 J and RT is 2493 J mol⁻¹, so a result close to one tenth of a mole is reasonable. This scale check can catch a thousandfold volume mistake without repeating every calculator entry.

If 2.50 is used directly as m³, the amount becomes about 100 mol rather than 0.100 mol. The algebra can be correct while the supplied sample has effectively been replaced by a volume one thousand times larger.

If 27 is substituted for 300, n becomes about 1.11 mol. Ask what temperature that input represents in the equation. The resulting molar mass would also change, showing how one early error propagates.

Use a reverse check: nRT/p should reconstruct approximately 0.00250 m³ using the retained amount. Reverse substitution tests internal consistency, while the original unit table checks whether the inputs still correspond to the intended sample.

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Give a new ideal, nonreacting mixture containing 0.0400 mol A and 0.0600 mol B in 2.50 dm³ at 300 K. Its total amount is 0.1000 mol. Do not identify it as exactly the same dataset as the rounded 100 kPa main example.

The total pressure is nRT/V = 0.1000 × 8.31 × 300/0.00250 = 99 720 Pa, or 99.7 kPa. The mole fractions are 0.400 for A and 0.600 for B.

Using Dalton’s law for the ideal mixture, pA = 0.400 × 99.720 = 39.9 kPa and pB = 0.600 × 99.720 = 59.8 kPa after rounding. Their unrounded sum equals the total pressure.

Each species occupies the whole container in this gas-mixture model. Do not assign A only 40% of the container volume and then combine that smaller volume with its partial pressure; that would change the intended relationship.

If the mixture reacts or forms a condensed phase, determine the remaining gaseous amounts before applying the relation. Partial pressure is linked to the actual gas-phase mole fraction, not necessarily the initial mixture fraction.

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For a supplied wet-gas extension, give total pressure 103 kPa and water-vapour partial pressure 3.00 kPa at the stated temperature. If the only gases are water vapour and the collected gas G, pG = 103 − 3.00 = 100 kPa.

Use that dry-gas partial pressure for the amount of G in the total gas volume. Do not use the full wet-mixture pressure and then claim the resulting amount belongs entirely to G. The species basis must remain consistent.

The ideal approximation is generally more useful at lower pressure and sufficiently high temperature away from condensation. At very high pressure, finite molecular size becomes important; at sufficiently low temperature, attractive interactions and condensation can matter.

Do not turn this qualitative trend into a universal numerical pressure threshold. Accuracy depends on the gas and conditions. The question should supply an ideal assumption or suitable evidence if a quantitative error bound is required.

For a density extension using the same main sample, density is 4.40/2.50 = 1.76 g dm⁻³. Combining density with pV = nRT gives M = ρRT/p with consistent units. This is the same sample information expressed through a different measured ratio.

With density 1.76 kg m⁻³, SI substitution gives M = 1.76 × 8.31 × 300/100 000 = 0.0438768 kg mol⁻¹, or 43.9 g mol⁻¹. The conversion from kilograms to grams is essential; the two numerical molar-mass values describe the same result in different units.

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If a gauge pressure is substituted without the atmospheric correction in the specified gauge example, ask what zero on that gauge means. The repair is the reference pressure, not a new gas equation.

If temperature is left in Celsius, ask whether doubling that number would double the absolute temperature. Use the kelvin conversion and the physical meaning of T rather than merely memorising an extra 273.

If volume is converted as though it were a length, ask how many cubic centimetres fit in a cubic metre. Reconstruct the cubed conversion factor, then return to the original sample.

If the molar mass is rounded too early from a rounded intermediate n, retain the amount through the final division. Distinguish harmless final rounding from a numerical drift caused by dropping useful digits prematurely.

If an ideal answer is presented as an exact measurement of a real gas, ask which assumptions have been tested. A correct calculation within a supplied model does not establish that the model describes every real condition exactly.

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Give a fresh ideal sample at 100 kPa absolute and 300 K with volume 5.00 dm³. Its amount is 100 000 × 0.00500/(8.31 × 300) = 0.20056 mol, reported as 0.201 mol.

The volume has doubled relative to the main example while p and T are unchanged, so the amount doubles. This proportional check helps establish whether the student understands the equation’s relationships as well as its numerical use.

Supply mass 8.80 g for this sample. Its molar mass is again 43.9 g mol⁻¹ using retained digits. The sample has more material but the same inferred molar mass under the stated ideal model.

Ask what happens to pressure if this fixed amount is placed in half the volume at the same temperature, with the ideal assumption retained. Pressure doubles. State which variables remain fixed before using the proportional reasoning.

For a parent update, describe whether the student can match the inputs, convert consistently and explain the scaling. Mention any continuing difficulty with pressure references or mixture amounts as a defined next teaching step.

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Check that pressure, volume, temperature and mass belong to the same stated sample and state. Verify absolute pressure, kelvin conversion, gas-constant units and any wet-gas correction independently from the arithmetic.

Keep total gas amount separate from the amount of one component. In a mixture task, check mole fractions, partial-pressure sums and whether any reaction or condensation changes the gaseous inventory.

Choose repair when unit conversion is insecure, stabilise through a scaled-volume comparison and extend to molar mass, density or partial pressure when the basic relation is independent. Each extension should introduce a clear new decision.

Audit identity claims and ideality language. An inferred molar mass is useful evidence, while unique identification and a quantitative statement of real-gas accuracy need additional support.

Retain the input table, one corrected calculation and a fresh independent result. They show whether the student can connect the gas equation to a consistent physical state rather than substitute unrelated numbers into a remembered formula.

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