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first law of thermodynamics questions with solutions
Practice Thermodyanmics for JEE Main, JEE Advanced, NEET, and Classes 11–12. Attempt each question, then open it for the full solution.
Showing 1–10 of 19 questions
- MCQmediumThermodyanmics
One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27°C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27°C doing the same amount of work W, then its final temperature will be (close to) __________ °C.
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JEE Mains 202623 January Evening Shift
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In the P–V diagram, a gas goes A(2V₀,P₀) → B(3V₀,P₀) → C(3V₀,2P₀) → D(2V₀,2P₀) → E(V₀,2P₀). Find the total work done by the gas from A to E.
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JEE Mains 2025 23 Jan Evening Shift
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Density of water at 4°C and 20°C are 1000 kg/m³ and 998 kg/m³ respectively. The increase in internal energy of 4 kg water when it is heated from 4°C to 20°C is____J.
(Specific heat capacity of water = 4.2 × 10³ J·kg⁻¹·°C⁻¹ and 1 atmospheric pressure = 10⁵ Pa)
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JEE Mains 202624 January Morning Shift
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Match List-I with List-II. List-I: (A) Isobaric, (B) Isochoric, (C) Adiabatic, (D) Isothermal. List-II: (I) ΔQ = ΔW, (II) ΔQ = ΔU, (III) ΔQ = 0, (IV) ΔQ = ΔU + PΔV. Choose the correct match.
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JEE Mains 2025 4 April Evening Shift
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For an ideal monoatomic gas undergoing an isobaric process, the ratio ΔQ/ΔU is:
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JEE Mains 2025 24 Jan Morning Shift
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10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from P₁ to P₂ is α Joule (P₁ = 21.7 Pa and P₂ = 30 Pa, Cv = 21 J/K·mol, R = 8.3 J/mol·K). The value of α is ____.
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JEE Mains 202624 January Evening Shift
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A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2°C. The work done by the gas is : (Given, R = 8.3 J K⁻¹ mol⁻¹.)
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JEE Mains 2024 06 Apr Evening Shift
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In an adiabatic process, which of the following statements is true?
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JEE Mains 2025 2 April Morning Shift
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The specific heat at constant pressure of a real gas obeying PV² = RT equation is :
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JEE Mains 2024 06 Apr Morning Shift
Open question & full solution → - NumericmediumThermodyanmics
As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition P₁ and a freely movable but thermally insulated piston P₂. The partition P₁ with thermal conductivity K, cross sectional area A and width x divides the container into two sections, S₁ and S₂, each containing one mole of a monoatomic gas. The piston P₂ moves freely such that the gas in S₂ is always at the atmospheric pressure. Initially, the difference between the temperatures of S₁ and S₂ is ΔT₀. The time it takes for the temperature difference to become ΔT₀/2 is nxR/(KA), where R is the universal gas constant. The value of n is: [Given: ln2 ≈ 0.7]
JEE Advanced2026Paper 1
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