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PhysicsMediumClass 11

Ideal Gas Law — PV = nRT, gas laws

Thermodynamics & KTG

21

JEE Qs

8%

Hard

75

min

Master unit conversions and always use Kelvin for temperature to avoid common errors in gas law problems.

🧮 Key Formulas

PV = nRT
P_1V_1/T_1 = P_2V_2/T_2
P = (rho)RT/M
PV = NkT
P_total = P_A + P_B + ...
R = 8.314 J/(mol·K) = 0.0821 L·atm/(mol·K) = 1.987 cal/(mol·K)
k = R/N_A = 1.38 x 10^-23 J/K

✅ Key Points for JEE

  • 1Always convert temperature to Kelvin (absolute temperature) before using any gas law equation.
  • 2Ensure consistency of units for pressure (P), volume (V), and the gas constant (R); choose the appropriate R value for the given units.
  • 3The ideal gas law is an equation of state that describes the macroscopic behavior of an ideal gas under various conditions.
  • 4The combined gas law (P₁V₁/T₁ = P₂V₂/T₂) is a powerful tool for relating initial and final states of a fixed amount of gas undergoing changes in P, V, or T.
  • 5Dalton's Law of Partial Pressures allows the calculation of the total pressure in a mixture of non-reacting ideal gases by summing the partial pressures of individual gases.

⚠️ Common Mistakes

  • Using temperature in Celsius (°C) instead of Kelvin (K) in gas law equations.
  • Inconsistent use of units for P, V, and R (e.g., using P in atm, V in m³, and R in J/mol·K).
  • Confusing 'n' (number of moles) with 'N' (number of molecules) and using the incorrect constant (R instead of k or vice versa).
  • Applying the ideal gas law to real gases at conditions of very high pressure or very low temperature, where ideal gas assumptions break down.

📝 Practice Questions

See all

Q29.Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process. Reason (R) : In isothermal process, PV = constant, while in adiabatic process PVγ = constant. Here γ is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas. In the light of the above statements, choose the correct answer from the options given below : (1) Both (A) and (R) are true and (R) is the correct (2) Both (A) and (R) are true but (R) is NOT the explanation of (A) correct explanation of (A) (3) (A) is true but (R) is false (4) (A) is false but (R) is true 2025 (29 Jan Shift 2) JEE Main Previous Year Paper

2025·Assertion ReasoningMedium

Q40.A cup of coffee cools from 90∘C to 80∘C in t minutes when the room temperature is 20∘C. The time taken by the similar cup of coffee to cool from 80∘C to 60∘C at the same room temperature is : (1) 13 10 t (2) 1013 t (3) 13 5 t (4) 135 t

2025·NumericalMedium

Q42. A poly-atomic molecule ( CV = 3R, CP = 4R, where R is gas constant) goes from phase space point A (PA = 105 Pa, VA = 4 × 10−6 m3) to point B (PB = 5 × 104 Pa, VB = 6 × 10−6 m3) to point 2025 (29 Jan Shift 2) JEE Main Previous Year Paper C (PC = 104 Pa, VC = 8 × 10−6 m3). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is : (1) 500R(ln 3 + ln 4) (2) 450R(ln 4 −ln 3) (3) 500R ln 2 (4) 400R ln 4

2025·MCQHard

Q29.An amount of ice of mass 10−3 kg and temperature −10∘C is transformed to vapour of temperature 110∘C by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice = 2100Jkg−1 K−1 , specific heat of water = 4180Jkg−1 K−1 , specific heat of steam = 1920Jkg−1 K−1 , Latent heat of ice = 3.35 × 105Jkg−1 and Latent heat of steam = 2.25 × 106 Jkg−1 ) (1) 3043 J (2) 3024 J (3) 3003 J (4) 3022 J

2025·MCQMedium

Q41.Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K . If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding temperature to be negligible), (1) 16 E (2) E (3) 64 E (4) 256 E

2025·MCQMedium

Q49.Three conductors of same length having thermal conductivity k1, k2 and k3 are connected as shown in figure. Area of cross sections of 1st and 2nd conductor are same and for 3rd conductor it is double of the 1st conductor. The temperatures are given in the figure. In steady state condition, the value of θ is _______ ∘C. (Given : k1 = 60Js−1 m−1 K−1, k2 = 120Js−1 m−1 K−1, k3 = 135Js−1 m−1 K−1 ) ∣ ∣∣ 2025 (22 Jan Shift 1) JEE Main Previous Year Paper

2025·NumericalMedium

NCERT Chapters

  • Class 11 Physics Ch 12: Thermodynamics
  • Class 11 Chemistry Ch 5: States of Matter