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Gaseous State flash cards

Master Gaseous State through 96 JEE Advanced-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Gaseous State, question and answer

19 of this chapter's 96 cards, laid out open so you can read straight through. The remaining 77 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.State the three common measurable properties used to describe a gas sample.

    Pressure PP, Volume VV, Temperature TT (and amount of gas nn). Gases have no fixed shape or volume, are highly compressible, exert equal pressure in all directions, and mix completely (diffuse) with one another.

    Hint: Four variables in PV=nRTPV=nRT.

  2. 2.State Boyle's law and give its mathematical form.

    At constant TT and nn, the volume of a fixed mass of gas is inversely proportional to its pressure: V1PV\propto \dfrac{1}{P}, i.e. PV=constantPV=\text{constant}, so P1V1=P2V2P_1V_1=P_2V_2.

    Hint: Squeeze harder, volume shrinks — TT fixed.

  3. 3.On a graph, what does a plot of PP vs 1V\dfrac{1}{V} look like for a gas obeying Boyle's law?

    A straight line passing through the origin with slope == constant =nRT=nRT. A plot of PP vs VV is a rectangular hyperbola (an isotherm).

    Hint: Linearise the inverse relation.

  4. 4.State Charles's law and its mathematical form.

    At constant PP and nn, volume is directly proportional to absolute temperature: VTV\propto T, so V1T1=V2T2\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2} (with TT in kelvin).

    Hint: Heat it, it expands — PP fixed.

  5. 5.State the volume-per-degree form of Charles's law and its significance for absolute zero.

    Vt=V0(1+t273.15)V_t=V_0\left(1+\dfrac{t}{273.15}\right). Extrapolating VV to zero gives t=273.15C=0Kt=-273.15\,^\circ\text{C}=0\,\text{K}, absolute zero, the temperature at which an ideal gas would have zero volume.

    Hint: Coefficient 1/273.151/273.15 per ^\circC.

  6. 6.State Gay-Lussac's (Amontons') law of pressure–temperature.

    At constant VV and nn, pressure is directly proportional to absolute temperature: PTP\propto T, so P1T1=P2T2\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}.

    Hint: Sealed rigid container heated.

  7. 7.State Avogadro's law.

    At the same TT and PP, equal volumes of all gases contain equal numbers of molecules: VnV\propto n (at constant T,PT,P). Equivalently V1n1=V2n2\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2}.

    Hint: Equal volumes ⇒ equal molecules.

  8. 8.What is the molar volume of an ideal gas at STP, and what are the current STP conditions?

    At STP (273.15K273.15\,\text{K}, 105Pa=1bar10^5\,\text{Pa}=1\,\text{bar}, IUPAC) molar volume =22.7L mol1=22.7\,\text{L mol}^{-1}. At the older STP (1atm1\,\text{atm}) it is 22.414L mol122.414\,\text{L mol}^{-1}.

    Hint: 22.7 L at 1 bar; 22.4 L at 1 atm.

  9. 9.Write the ideal gas equation and identify each term.

    PV=nRTPV=nRT, where PP = pressure, VV = volume, nn = moles, TT = absolute temperature, RR = universal gas constant. It combines Boyle's, Charles's and Avogadro's laws.

    Hint: The master equation.

  10. 10.Give the value of the gas constant RR in three common sets of units.

    R=0.0821L atm K1mol1=8.314J K1mol1=2cal K1mol1R=0.0821\,\text{L atm K}^{-1}\text{mol}^{-1}=8.314\,\text{J K}^{-1}\text{mol}^{-1}=2\,\text{cal K}^{-1}\text{mol}^{-1} (approx). Also 8.314×107erg K1mol18.314\times10^{7}\,\text{erg K}^{-1}\text{mol}^{-1}.

    Hint: 0.0821, 8.314, ~2.

  11. 11.How is the ideal gas equation rewritten in terms of density dd and molar mass MM?

    Since n=wMn=\dfrac{w}{M} and d=wVd=\dfrac{w}{V}: PM=dRTPM=dRT, so M=dRTPM=\dfrac{dRT}{P}. Density is directly proportional to pressure and molar mass, inversely to temperature.

    Hint: Replace nn with w/Mw/M.

  12. 12.What is the combined gas law?

    For a fixed amount of gas: P1V1T1=P2V2T2\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}. It merges Boyle's, Charles's and Gay-Lussac's laws for changes in state.

    Hint: PV/TPV/T constant.

  13. 13.Distinguish an ideal gas from a real gas conceptually.

    An ideal gas obeys PV=nRTPV=nRT at all P,TP,T; its molecules have negligible volume and no intermolecular forces. Real gases have finite molecular size and attractive/repulsive forces, so they obey the ideal law only at low pressure and high temperature.

    Hint: No size, no forces = ideal.

  14. 14.State Dalton's law of partial pressures.

    The total pressure of a mixture of non-reacting gases equals the sum of the partial pressures of the components: Ptotal=P1+P2+P3+P_\text{total}=P_1+P_2+P_3+\cdots. Each partial pressure is the pressure the gas would exert alone in the same volume.

    Hint: Pressures add up.

  15. 15.Relate the partial pressure of a component to its mole fraction.

    Pi=xiPtotalP_i=x_i\,P_\text{total}, where xi=nintotalx_i=\dfrac{n_i}{n_\text{total}} is the mole fraction of component ii. Partial pressure is total pressure times mole fraction.

    Hint: Pi=xiPTP_i = x_i P_T.

  16. 16.How is the pressure of a gas collected over water corrected?

    Pdry gas=PtotalpH2OP_\text{dry gas}=P_\text{total}-p_{\text{H}_2\text{O}}, where pH2Op_{\text{H}_2\text{O}} (aqueous tension) is the saturated vapour pressure of water at that temperature. Subtract the water vapour pressure.

    Hint: Subtract aqueous tension.

  17. 17.State Graham's law of diffusion/effusion.

    At the same TT and PP, the rate of diffusion (or effusion) is inversely proportional to the square root of the density (or molar mass): r1d1Mr\propto\dfrac{1}{\sqrt{d}}\propto\dfrac{1}{\sqrt{M}}, so r1r2=M2M1\dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1}}.

    Hint: Lighter gas diffuses faster.

  18. 18.Distinguish diffusion from effusion.

    Diffusion is the spontaneous intermingling of gases due to random molecular motion. Effusion is the escape of gas molecules through a tiny hole into vacuum/low pressure. Both follow Graham's law dependence on M\sqrt{M}.

    Hint: Mixing vs escaping through a pinhole.

  19. 19.State the fundamental postulates of the kinetic theory of gases.

    Gas consists of tiny particles in constant random motion; the volume of particles is negligible vs container volume; there are no intermolecular forces; collisions are perfectly elastic (no KE loss); average kinetic energy is directly proportional to absolute temperature.

    Hint: Point masses, elastic, KE ∝ T.

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