Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
A mixture of carbon dioxide and oxygen has volume 8310 cm$^3$, temperature 300 K, pressure 100 kPa and mass 13.2 g. The number of moles of carbon dioxide and oxygen gases in the mixture respectively are ________. (Assume both carbon dioxide and oxygen gases behave like ideal gases) [R = 8.31 J/mol.K]
(1) 0.15 and 0.18
(2) 0.25 and 0.08
(3) 0.21 and 0.12
(4) 0.13 and 0.20
Show answer and solution
Answer: Option (3)
Key idea: Ideal Gas Law with Gas Mixtures - Using PV=nRT and mass constraints to find individual moles of gases in a mixture
Convert volume to SI units: $V = 8310 ext{ cm}^3 = 8.31 imes 10^{-3} ext{ m}^3$
Use ideal gas equation to find total moles: $n_{total} = \frac{PV}{RT} = \frac{100000 imes 8.31 imes 10^{-3}}{8.31 imes 300} = \frac{1}{3}$ mol
Set up equations with $n_1$ (CO₂) and $n_2$ (O₂): $n_1 + n_2 = 0.333$ and $44n_1 + 32n_2 = 13.2$
Substitute $n_2 = 0.333 - n_1$ into mass equation: $44n_1 + 32(0.333 - n_1) = 13.2$
Solve for $n_1$: $12n_1 = 13.2 - 10.67 = 2.53$, so $n_1 = 0.21$ mol
One gas of $n_1$ mole of molecules at temperature $T_1$, volume $V_1$, and pressure $P_1$, and another gas of $n_2$ mole of molecules at temperature $T_2$, volume $V_2$, and pressure $P_2$, are mixed resulting in pressure $P$ and volume $V$ of the mixture. The temperature of the mixture is _______.
(1) $(T_1 + T_2)/2$
(2) $T_1 T_2 PV/(T_2 P_1 V_1 + T_1 P_2 V_2)$
(3) $(T_2 P_1 V_1 + T_1 P_2 V_2)/(T_1 T_2 PV)$
(4) $|T_1 - T_2|/2$
Show answer and solution
Answer: Option (2)
Key idea: Ideal Gas Law and Gas Mixing
Apply ideal gas law to gas 1: $P_1V_1 = n_1RT_1$, so $n_1 = \frac{P_1V_1}{RT_1}$
Apply ideal gas law to gas 2: $P_2V_2 = n_2RT_2$, so $n_2 = \frac{P_2V_2}{RT_2}$
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A: If the average kinetic energy of H$_2$ and O$_2$ molecules, kept in two different sized containers are same, then their temperatures will be same.
Reason R: The r.m.s. speed of H$_2$ and O$_2$ molecules are same at same temperature.
Choose the *correct* answer from the options given below
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
Show answer and solution
Answer: Option (3)
Key idea: Kinetic Theory: Average Kinetic Energy and RMS Speed
**Check Assertion A:** Average kinetic energy is $KE_{avg} = \frac{3}{2}kT$, which depends only on temperature $T$, not on molecular mass or container size
If $KE_{avg}(H_2) = KE_{avg}(O_2)$, then $\frac{3}{2}kT_1 = \frac{3}{2}kT_2$, giving $T_1 = T_2$
**Assertion A is TRUE**: Same average KE means same temperature
**Check Reason R:** RMS speed is $v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3kT}{m}}$, which depends on both temperature AND molecular mass
At same temperature: $\frac{v_{rms}(H_2)}{v_{rms}(O_2)} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32}{2}} = 4$
**Reason R is FALSE**: RMS speeds are different at same temperature (lighter molecules move faster)
**Conclusion:** A is true but R is false → Option (3)
An ideal gas at pressure $P$ and temperature $T$ is expanding such that $PT^3 = constant$. The coefficient of volume expansion of the gas is __________.
(1) $\frac{2}{T}$
(2) $\frac{1}{T}$
(3)
(4)
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Coefficient of volume expansion for ideal gas in a polytropic process
Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure $90$ kPa and temperature $400$ K. Keeping the temperature of one vessel constant at $400$ K the second vessel temperature is raised to $500$ K. The final pressure in the vessels is ______ kPa.
(1) 100
(2) 120
(3) 90
(4) 105
Show answer and solution
Answer: Option (1)
Key idea: Conservation of moles in connected vessels with different temperatures using ideal gas law
Calculate initial moles in each vessel: $n_1 = n_2 = \frac{P_1V}{RT_1} = \frac{90V}{400R}$
Find total moles: $n_{total} = 2 \times \frac{90V}{400R} = \frac{9V}{20R}$
After heating, final pressure $P'$ is same in both vessels. Write moles in each vessel: $n_1' = \frac{P'V}{400R}$ and $n_2' = \frac{P'V}{500R}$
Apply conservation of moles: $\frac{P'V}{400R} + \frac{P'V}{500R} = \frac{9V}{20R}$
If 2 mole of an ideal monoatomic gas at temperature $T$, is mixed with 6 mole of another ideal monoatomic gas at temperature $2T$ then the temperature of mixture is :
(1) $\frac{5}{2}T$
(2) $\frac{5}{4}T$
(3) $\frac{7}{2}T$
(4) $\frac{7}{4}T$
Show answer and solution
Answer: Option (4)
Key idea: Conservation of Internal Energy in Gas Mixtures
For monoatomic gas, $C_v = \frac{3}{2}R$ and internal energy $U = nC_vT$
Calculate initial internal energy of gas 1: $U_1 = 2 \times \frac{3}{2}R \times T = 3RT$
Calculate initial internal energy of gas 2: $U_2 = 6 \times \frac{3}{2}R \times 2T = 18RT$
Total initial energy: $U_{initial} = 3RT + 18RT = 21RT$
After mixing, total moles = 8, so $U_{final} = 8 \times \frac{3}{2}R \times T_f = 12RT_f$
By energy conservation: $21RT = 12RT_f$, giving $T_f = \frac{7T}{4}$
FAQs
Frequently asked questions
How many questions come from Kinetic Theory of Gases in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 6 questions on it, about 0.7 per shift.
Where can I practise Kinetic Theory of Gases PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
