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  2. MCAT Chemical and Physical Foundations of Biological Systems
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MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 5e Thermodynamics Energy Changes

Study 5e Thermodynamics Energy Changes in MCAT Chemical and Physical Foundations of Biological Systems with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on 5e Thermodynamics Energy Changes, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 5e Thermodynamics Energy Changes

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QUESTION

Which sign of ΔH\Delta HΔH indicates an exothermic reaction at constant pressure?

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ANSWER

ΔH<0\Delta H<0ΔH<0. Exothermic reactions release heat, corresponding to a decrease in the system's enthalpy.

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Flashcard 1: Which sign of ΔH\Delta HΔH indicates an exothermic reaction at constant pressure?

Answer: ΔH<0\Delta H<0ΔH<0. Exothermic reactions release heat, corresponding to a decrease in the system's enthalpy.

Flashcard 2: What is the sign convention for qqq when heat enters the system?

Answer: q>0q>0q>0. In thermodynamic conventions, heat absorbed by the system from surroundings is assigned a positive value.

Flashcard 3: What is the sign convention for www when the system does work on the surroundings?

Answer: w<0w<0w<0. Work performed by the system on surroundings reduces the system's energy, thus assigned a negative sign.

Flashcard 4: State the first law of thermodynamics using ΔU\Delta UΔU, qqq, and www.

Answer: ΔU=q+w\Delta U=q+wΔU=q+w. The first law states that internal energy change equals heat added to the system plus work done on the system.

Flashcard 5: What is the formula for pressure–volume work at constant external pressure?

Answer: w=−PextΔVw=-P_{\text{ext}}\Delta Vw=−Pext​ΔV. For irreversible expansion at constant external pressure, work equals the negative product of pressure and volume change.

Flashcard 6: For an isochoric process, what is the work www?

Answer: w=0w=0w=0. Isochoric processes involve no volume change, hence no pressure-volume work is performed.

Flashcard 7: For an adiabatic process, what is the heat transfer qqq?

Answer: q=0q=0q=0. Adiabatic processes, by definition, prohibit heat exchange between the system and surroundings.

Flashcard 8: What is the relationship between ΔH\Delta HΔH and qqq at constant pressure (PV work only)?

Answer: ΔH=qp\Delta H=q_pΔH=qp​. Enthalpy change equals heat transferred at constant pressure when only PV work occurs.

Flashcard 9: What is the relationship between ΔU\Delta UΔU and qqq at constant volume?

Answer: ΔU=qv\Delta U=q_vΔU=qv​. At constant volume, internal energy change equals heat transferred since no work is done.

Flashcard 10: State the definition of enthalpy in terms of UUU, PPP, and VVV.

Answer: H=U+PVH=U+PVH=U+PV. Enthalpy accounts for internal energy plus the energy associated with pressure-volume work.

Flashcard 11: State the ideal gas law relating PPP, VVV, nnn, and TTT.

Answer: PV=nRTPV=nRTPV=nRT. The ideal gas law describes the proportional relationship between pressure, volume, moles, and temperature.

Flashcard 12: What is the formula for heat absorbed at constant pressure for a temperature change ΔT\Delta TΔT?

Answer: qp=nCpΔTq_p=nC_p\Delta Tqp​=nCp​ΔT. Heat capacity at constant pressure quantifies energy required to raise temperature for a given amount of substance.

Flashcard 13: What is the formula for heat absorbed at constant volume for a temperature change ΔT\Delta TΔT?

Answer: qv=nCvΔTq_v=nC_v\Delta Tqv​=nCv​ΔT. Heat capacity at constant volume measures energy input for temperature change without volume alteration.

Flashcard 14: State Hess's law for combining reaction enthalpies.

Answer: ΔHnet=∑ΔHsteps\Delta H_{\text{net}}=\sum \Delta H_{\text{steps}}ΔHnet​=∑ΔHsteps​. Since enthalpy is a state function, the net change equals the sum of stepwise changes regardless of path.

Flashcard 15: State the definition of entropy change for a reversible process.

Answer: ΔS=qrevT\Delta S=\frac{q_{\text{rev}}}{T}ΔS=Tqrev​​. For reversible processes at constant temperature, entropy change is reversible heat divided by temperature.

Flashcard 16: What is the criterion for spontaneity in terms of the universe entropy change?

Answer: Spontaneous if ΔSuniv>0\Delta S_{\text{univ}}>0ΔSuniv​>0. The second law dictates that spontaneous processes increase the total entropy of the universe.

Flashcard 17: State the Gibbs free energy equation relating ΔG\Delta GΔG, ΔH\Delta HΔH, TTT, and ΔS\Delta SΔS.

Answer: ΔG=ΔH−TΔS\Delta G=\Delta H-T\Delta SΔG=ΔH−TΔS. Gibbs free energy combines enthalpy and entropy to assess spontaneity at constant temperature and pressure.

Flashcard 18: At constant TTT and PPP, what sign of ΔG\Delta GΔG indicates a spontaneous process?

Answer: ΔG<0\Delta G<0ΔG<0. Negative Gibbs free energy change indicates the process can occur spontaneously under those conditions.

Flashcard 19: What is the relationship between ΔG∘\Delta G^\circΔG∘ and the equilibrium constant KKK?

Answer: ΔG∘=−RTln⁡K\Delta G^\circ=-RT\ln KΔG∘=−RTlnK. This equation links thermodynamic favorability under standard conditions to the equilibrium position.

Flashcard 20: Find ΔU\Delta UΔU if q=−50 Jq=-50\,\text{J}q=−50J and w=+20 Jw=+20\,\text{J}w=+20J for the system.

Answer: ΔU=−30 J\Delta U=-30\,\text{J}ΔU=−30J. The first law calculates internal energy change as the sum of heat and work with proper signs.

Flashcard 21: Find www if Pext=2 atmP_{\text{ext}}=2\,\text{atm}Pext​=2atm and the system expands by ΔV=3 L\Delta V=3\,\text{L}ΔV=3L.

Answer: w=−6 L\cdotpatmw=-6\,\text{L·atm}w=−6L\cdotpatm. Expansion work against constant pressure is negative the product of pressure and volume increase.

Flashcard 22: Find qpq_pqp​ if n=2n=2n=2 and Cp=30 J\cdotpmol−1\cdotpK−1C_p=30\,\text{J·mol}^{-1}\text{·K}^{-1}Cp​=30J\cdotpmol−1\cdotpK−1 for ΔT=10 K\Delta T=10\,\text{K}ΔT=10K.

Answer: qp=600 Jq_p=600\,\text{J}qp​=600J. Constant pressure heat transfer uses molar heat capacity to relate energy to temperature change.

Flashcard 23: Find ΔG\Delta GΔG at T=300 KT=300\,\text{K}T=300K if ΔH=20 kJ\Delta H=20\,\text{kJ}ΔH=20kJ and ΔS=50 J\cdotpK−1\Delta S=50\,\text{J·K}^{-1}ΔS=50J\cdotpK−1.

Answer: ΔG=5 kJ\Delta G=5\,\text{kJ}ΔG=5kJ. Gibbs free energy subtracts the entropy term from enthalpy, ensuring unit consistency in calculations.

Flashcard 24: What is the relationship between CpC_pCp​ and CvC_vCv​ for an ideal gas?

Answer: Cp=Cv+RC_p=C_v+RCp​=Cv​+R. The difference arises from the additional work term at constant pressure for ideal gases.