Physics · Kinetic Theory · NEET
Both gases fill the whole volume V and share the same temperature T. Each gas pushes on its own using its own moles: partial pressure of gas 1 is P1 = n1RT/V and of gas 2 is P2 = n2RT/V. By Dalton's law the total pressure is just P = P1 + P2 = (n1 + n2)RT/V. So you only need the total number of moles. The kind of gas does not matter for pressure, only how many moles there are.
You do not add temperatures. Temperature is not a stored amount you can pile up. If both gases are already at the same temperature T, the mixture stays at T. If they start at different temperatures, they exchange heat until they reach one common final temperature, which you find from energy balance, not by adding. In almost every NEET problem the two gases are given at the same T, so you simply use that T for both.
Pressure depends only on how many molecules hit the walls, so every mole counts the same and you add moles. Internal energy depends on the degrees of freedom f of each gas, because energy is stored in motion (translation, rotation). A monatomic gas stores less energy per mole (f = 3) than a diatomic gas (f = 5). So you must compute each gas's energy separately with its own f and then add: U = (f1/2)n1RT + (f2/2)n2RT.
Monatomic gases (He, Ne, Ar) have f = 3 (only translation). Diatomic gases at ordinary temperature (O2, N2, H2) have f = 5 (3 translation + 2 rotation), if you neglect vibration. This is the key number that decides the energy of each gas in the mixture. NEET often says 'neglect vibrational modes' to tell you to use f = 5 for diatomic, not 7.
If temperature stays the same, use PV = (moles)RT. The product PV measures the total gas energy, and it is conserved when you just open a wall (no heat added, temperature fixed). So final pressure P = (P1V1 + P2V2) / (V1 + V2). This is a weighted average by volume. NEET 2025 asked exactly this: 1 atm in 2 L and 2 atm in 3 L give P = (1x2 + 2x3)/(2+3) = 8/5 = 1.6 atm.
A gas mixture consists of 2 moles of O2 and 4 moles of Ar at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:
A container of volume 200 cm3 contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The pressure of the system at temperature 200 K (R = 8.3 J K-1 mol-1) will be:
A container has two chambers of volumes V1 = 2 litres and V2 = 3 litres separated by a partition. The chambers contain n1 = 5 and n2 = 4 moles of ideal gas at pressures p1 = 1 atm and p2 = 2 atm respectively. When the partition is removed, the mixture attains an equilibrium pressure of:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
P = (n1 + n2)RT / V, which is the same as adding the partial pressures P1 + P2. This is Dalton's law: each gas contributes pressure using its own moles, and you add them.
Add each gas's energy using its own degrees of freedom: U = (f1/2)n1RT + (f2/2)n2RT. Use f = 3 for monatomic gases and f = 5 for diatomic gases (neglecting vibration).
No. If both gases are already at the same temperature T, the mixture stays at T. Temperature is not added; only moles and energies are handled with their rules.
Ideal gas molecules do not interact, so each gas behaves as if the other is not there. The pressures simply add, and the energies simply add. This lets you treat the mixture as one gas for the equation of state.
Oxygen (O2) is diatomic, so f = 5 (3 translational + 2 rotational) at ordinary temperature when vibration is neglected. This gives its internal energy as (5/2)nRT.
