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Q18. First Law of Thermodynamics

Q18. First Law of Thermodynamics. A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal or isobaric. Rank those three processes in order of the work required of an external agent, least to greatest.

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Q18. First Law of Thermodynamics

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  1. Q18. First Law of Thermodynamics

  2. A quantity of an ideal gas is compressed to half its initial volume. The process may be adiabatic, isothermal or isobaric. Rank those three processes in order of the work required of an external agent, least to greatest. adiabatic, isothermal, isobaric adiabatic, isobaric, isothermal isothermal, adiabatic, isobaric isobaric, adiabatic, isothermal isobaric, isothermal, adiabatic

  3. 2 P Adiabatic Isothermal Isobaric 2P P

  4. When an ideal gas undergoes a slow isothermal expansion: • (work done by the gas) = (energy absorbed as heat ) • (work done by the environment ) = (energy absorbed as heat ) • (increase in internal energy ) = ( heat absorbed ) • (increase in internal energy ) = ( work done by the gas ) • ( increase in internal energy ) = ( work done by the environment )

  5. Isothermal  dU = 0  d Q = d W Expansion  d W = P d V > 0 (Work done by gas) d Q > 0  Heat absorbed • (work done by the gas) = (energy absorbed as heat ) d W = d Q • (work done by the environment ) = (energy absorbed as heat ) d W = d Q • (increase in internal energy ) = ( heat absorbed ) d U = d Q • (increase in internal energy ) = ( work done by the gas ) d U = d W • ( increase in internal energy ) = ( work done by the environment ) d U =  d W

  6. An ideal gas of N monatomic molecules is in thermal equilibrium with an ideal gas of the same number of diatomic molecules and equilibrium is maintained as temperature is increased. The ratio of the changes in the internal energies ΔEdia / Emon is : 1/2 3/5 1 5/3 2

  7. The pressure of an ideal gas of diatomic molecules is doubled by halving the volume. The ratio of the new internal energy to the old, both measured relative to the internal energy at 0 K, is: 1/4 1/2 1 2 4

  8. T remains the same. So does U.

  9. When work W is done on an ideal gas of N diatomic molecules in thermal isolation the temperature increases by W/2Nk W/3Nk 2W/3Nk 2W/5Nk W/Nk

  10. Work W done on gas: Thermal isolation : dQ = 0  Diatomic molecules :

  11. When work W is done on an ideal gas of diatomic molecules in thermal isolation the increase in the total rotational energy of the molecules is: 0 W/3 2W/3 2W/5 W

  12. Work W done on gas: Thermal isolation : dQ = 0  Diatomic molecules :

  13. The pressure of an ideal gas is doubled during a process in which the energy given up as heat by the gas equals the work done on the gas. As a result, the volume is: doubled halved unchanged need more information to answer nonsense, the process is impossible

  14. Heat Q given up by the gas equals work W done on the gas    

  15. The temperature of n moles of an ideal monatomic gas is increased by T at constant pressure. The energy Q absorbed as heat, change Eint in internal energy, and work W done by the environment are given by: Q = (5/2)nRDT,DEint = 0, W = –nRDT Q = (3/2)nRDT,DEint = (5/2)nRDT,W = –(3/2)nRDT Q = (5/2)nRDT,DEint = (3/2)nRDT,W = nRDT Q = (3/2)nRDT,DEint = 0, W = nRDT Q = (5/2)nRDT,DEint = (3/2)nRDT,W = –nRDT

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