Exams › JEE Main › Physics

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by ΔT. The amount of heat absorbed by gas is (R is gas constant):

1. 1/2 K R ΔT
2. 1/2 R ΔT
3. 3/2 R ΔT
4. 2K/3 ΔT

Correct answer: 1/2 R ΔT

Solution

For 1 mole monoatomic, dU = (3/2)R*dT. With V=K/T, PV=RT gives P=RT^2/K, and W = integral P dV = -R*dT. So Q = (3/2)R*dT - R*dT = (1/2)R*dT.

Related JEE Main Physics questions

• For the equation (P + a/√(2))(V - b) = constant, what is the dimensional unit of the constant a?
• A refrigerator operates between 4°C and 30°C. To maintain its interior at a steady temperature, it must extract 600 calories of heat each second. If 1 calorie = 4.2 joule, the power needed is:
• A diatomic gas with adiabatic index γ = 1.4 is initially at a pressure of 2 atm. It is compressed adiabatically until its temperature increases from 27°C to 927°C. What is the final pressure of the gas?
• A Carnot engine operates with a diatomic ideal gas as its working fluid. During the adiabatic expansion stage, the gas volume rises from V to 32V. What is the efficiency of the engine?
• A spherical cavity of radius R contains black-body radiation at temperature T. Treat the radiation as a photon gas with energy density u = U/V proportional to T⁴ and pressure p = (1/3)(U/V). If the cavity expands adiabatically, how are T and R related?
• A metal has low-temperature specific heat capacity given by C = 32 (T/400)³ J K^−1 kg^−1. If a 100 g container made of this metal is to be cooled from 20 K down to 4 K using a special refrigerator working at room temperature (27°C), the work needed for the cooling process is

⚔️ Practice JEE Main Physics free + battle 1v1 →