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An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature. A. The work done by gas during the process is zero. B. The heat added to gas is different from change in its internal energy. C. The volume of the gas is increased. D. The internal energy of the gas is increased. E. The process is isochoric (constant volume process) Choose the correct answer from the options given below :-

Solution & Explanation

### Related Formula From the Ideal Gas Law: PV = nRT According to the First Law of Thermodynamics: Delta Q = Delta U + W ### Core Logic The question states that pressure increases linearly with temperature, which means their ratio is constant : P = kT implies fracPT = textconstant Since fracPT = fracnRV, the volume V must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true). ### Step 1: Evaluate All Statements * Statement A: True. In an isochoric process, dV = 0 implies W = int P dV = 0. * Statement B: False. Since work is zero, the First Law simplifies to Delta Q = Delta U, meaning heat added equals the change in internal energy. * Statement C: False. Volume is constant, so it does not increase. * Statement D: True. As pressure increases linearly with temperature, temperature increases, which causes the internal energy of the gas to increase. ### Step 2: Final Selection Gathering the true statements (A, D, and E) points directly to Option (2). ### Pattern Recognition A linear P-T line passing through the origin always indicates a constant volume graph. For constant volume graphs, work done is zero, which simplifies the first law calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

Reference Study Guides

More Thermodynamics Previous-Year Questions — Page 6

Q47 2025 Hess's Law / Enthalpy of Formation
Consider the following data: Heat of formation of CO_2(g) = -393.5mathrm~kJ~mol^-1 Heat of formation of H_2O(l) = -286.0mathrm~kJ~mol^-1 Heat of combustion of benzene = -3267.0mathrm~kJ~mol^-1 The heat of formation of benzene is ______ mathrmkJ~mol^-1 (Nearest integer).
Numerical Answer. Answer: 48 to 48

Solution

### Related Formula Enthalpy of reaction from enthalpy of formation data: Delta H_textreaction = sum Delta H_f(textProducts) - sum Delta H_f(textReactants) ### Core Logic Write out the balanced thermochemical equation for the combustion of benzene (C_6H_6): C_6H_6(l) + frac152O_2(g) rightarrow 6CO_2(g) + 3H_2O(l) Given parameters: - Delta H_c = -3267.0mathrm\ kJ/mol - Delta H_f[CO_2] = -393.5mathrm\ kJ/mol - Delta H_f[H_2O] = -286.0mathrm\ kJ/mol - Delta H_f[O_2] = 0mathrm\ kJ/mol ### Step 1: Applying Hess's Law Substitute these values into the reaction expression: -3267 = [6(-393.5) + 3(-286.0)] - Delta H_f[C_6H_6] -3267 = [-2361.0 - 858.0] - Delta H_f[C_6H_6] -3267 = -3219.0 - Delta H_f[C_6H_6] Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48mathrm\ kJ/mol ### Pattern Recognition Always set up products minus reactants when using heat of formation data. Pay close attention to stoichiometric coefficients (multiply CO_2 by 6 and H_2O by 3) to ensure accurate bookkeeping. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics
Q19 2025 Adiabatic Process
The workdone in an adiabatic change in an ideal gas depends upon only : [cite: 1, 2]
  • A. change in its pressure
  • B. change in its specific heat
  • C. change in its volume
  • D. change in its temperature

Solution

### Related Formula Delta W = -Delta U = -n C_v Delta T ### Core Logic In an adiabatic system, no heat exchange occurs (Q=0). By the first law of thermodynamics, Delta W = -Delta U. Since internal energy U depends explicitly on temperature metrics, the total work output shifts uniquely based on temperature variation Delta T. ### Chapter Mix Class 11 Physics: Thermodynamics

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