A stainless steel chamber contains Ar gas at a temperature T and pressure P. The total number of Ar atoms in the chamber is n. Now Ar gas in the chamber is replaced by CO₂ gas and the total number of CO₂ molecules in the chamber is n/2 at the same temperature T. The pressure in the chamber now is P′. Which one of the following relations holds true? (Both the gases behave as ideal gases)
- (a)P′ = P
- (b)P′ = 2P
- (c)P′ = P/2
- (d)P′ = P/4
Answer
Why
Correct: C, P′ = P/2. Write the ideal gas law in the form that counts particles rather than moles: PV = NkT, where N is the number of atoms or molecules and k is Boltzmann's constant. The chamber is a rigid stainless-steel vessel, so V does not change, and the question holds the temperature at T throughout. With V and T both fixed, pressure is proportional to N and to nothing else. Replacing n argon atoms by n/2 carbon dioxide molecules halves the particle count, so it halves the pressure: P′ = P/2. The identity of the gas never enters. That the CO₂ molecule is heavier than the argon atom, and triatomic rather than monatomic, is exactly the irrelevance the question is planted to test: Avogadro's principle says equal numbers of particles at the same volume and temperature exert the same pressure whatever they are made of.
Why the others are wrong
- (a)P′ = P: This would require the number of particles to be unchanged. It has been halved. A student reaches this answer by reasoning that the heavier CO₂ molecules hit the walls harder and make up for there being fewer of them, but at the same temperature the two effects cancel exactly, which is what makes pressure independent of molar mass.
- (b)P′ = 2P: The wrong direction altogether. Fewer particles in the same volume at the same temperature means fewer collisions with the walls each second, so the pressure must fall, not rise.
- (d)P′ = P/4: Halving applied twice: once correctly for the particle count and once again for some supposed effect of the change of gas. The molar mass of the gas does not appear anywhere in PV = NkT, so there is no second factor to apply.
Concept
The ideal gas law can be written PV = nRT, counting in moles, or PV = NkT, counting individual particles, with k = R ÷ N_A linking the two. Either form says the same thing: the pressure a gas exerts depends on how many particles are present, how much room they have and how hot they are, and on nothing else. The kinetic-theory picture behind it is that pressure is the average effect of particles striking the walls, and at a given temperature every gas has the same average translational kinetic energy per particle.
The examiner's technique here is to load the stem with information that does not matter (the vessel is stainless steel, the gas changes from a monatomic noble gas to a triatomic molecular one) and then hide the single relevant change in the middle. Strip the problem down before calculating: what is held constant, what actually changes? Here V is constant because the chamber is rigid, T is constant because the question says so, and N goes from n to n/2. With two of the three variables pinned, the ratio follows in one line. Heavier molecules do move more slowly at a given temperature, but they carry more momentum per collision, and the two effects cancel precisely, which is the physical content of Avogadro's principle.
Key facts
- The ideal gas law is PV = nRT in moles, or PV = NkT in particles; at fixed V and T, pressure is proportional to the number of particles.
- By Avogadro's principle, equal volumes of any ideal gases at the same temperature and pressure contain equal numbers of particles.
- Pressure does not depend on the molar mass of the gas: heavier particles move more slowly but strike harder, and the effects cancel.
- Boltzmann's constant k equals R ÷ N_A, about 1·38 × 10⁻²³ J K⁻¹.
- A rigid container means an isochoric change, in which volume is held constant and pressure and temperature vary together.
The switch from argon to carbon dioxide is a red herring: option (c).
Study next
Common traps
- Bringing the molar mass of the gas into a pressure calculation, where it has no place.
- Missing that a rigid metal chamber means the volume is fixed.
- Halving twice and arriving at P/4.
NDA sets the ideal gas law as a ratio problem (one or two quantities changed, the rest held fixed) with an irrelevant change of gas or of container material planted in the stem.
Related PYQs
If we plot a graph between volume V and inverse of pressure P (i.e., 1/P) for an ideal gas at constant temperature T, the curve so obtained is
- (a) straight line
- (b) circle
- (c) parabola
- (d) hyperbola
Answer(a) straight line
The same equation approached graphically, and a reminder that these items are all solved by fixing which variables the problem holds constant.
Practice
- practice, not a real PYQ
A rigid vessel contains an ideal gas at pressure P and temperature T. If the absolute temperature is doubled while the number of molecules and the volume stay the same, the new pressure is
- (a)P/2
- (b)P
- (c)2P
- (d)4P
Answer(c) 2P: at fixed volume and particle number, PV = NkT makes pressure proportional to absolute temperature. - practice, not a real PYQ
Two identical rigid flasks at the same temperature contain equal numbers of molecules, one of hydrogen and one of oxygen. The pressures in the two flasks are
- (a)equal
- (b)greater in the hydrogen flask
- (c)greater in the oxygen flask
- (d)in the ratio 1 : 16
Answer(a) equal: by Avogadro's principle, pressure depends on the number of particles and not on their mass.