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A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800~mathrmcm^3 and temperature 27^circmathrmC . The change in temperature when the gas is adiabatically compressed to 200~mathrmcm^3 is: (Take gamma = 1.5)

Solution & Explanation

### Related Formula For an adiabatic process: T V^gamma - 1 = textconstant where, T = absolute temperature in Kelvin, V = volume of the gas, gamma = adiabatic exponent. ### Core Logic Given values: - Initial volume, V_1 = 800mathrm~cm^3 - Final volume, V_2 = 200mathrm~cm^3 - Initial temperature, T_1 = 27^circmathrmC = 27 + 273 = 300mathrm~K - Adiabatic exponent, gamma = 1.5 implies gamma - 1 = 0.5 ### Step 1: Calculating Final Temperature Apply the adiabatic relation: T_1 V_1^gamma - 1 = T_2 V_2^gamma - 1 T_2 = T_1 left(fracV_1V_2right)^gamma - 1 Substitute the values: T_2 = 300 left(frac800200right)^0.5 = 300 times (4)^0.5 T_2 = 300 times 2 = 600mathrm~K ### Step 2: Calculating Change in Temperature Now compute the change in temperature (Delta T): Delta T = T_2 - T_1 = 600mathrm~K - 300mathrm~K = 300mathrm~K ### Pattern Recognition Always read carefully to see if the question asks for the **final temperature** or the **change in temperature**. Many students lose marks by choosing 600mathrm~K (the final temperature) instead of the difference 300mathrm~K! Stay sharp. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermodynamics

Reference Study Guides

Thermodynamics

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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