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.

MCAT Chemical and Physical Foundations of Biological Systems

5e Thermodynamics Energy Changes

0 mastered0 still learning

0% Complete

QUESTION
1/ 24

State the definition of entropy change for a reversible process.

Tap card or press Space to flip

ANSWER

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

How well did you know it?

Card 1 / 24

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.

All flashcards

Flashcard 1: State the definition of entropy change for a reversible process.

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

Flashcard 2: For an adiabatic process, what is the heat transfer qq?

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

Flashcard 3: At constant TT and PP, what sign of ΔG\Delta G indicates a spontaneous process?

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

Flashcard 4: State the definition of enthalpy in terms of UU, PP, and VV.

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

Flashcard 5: What is the formula for heat absorbed at constant pressure for a temperature change ΔT\Delta T?

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

Flashcard 6: Find ΔU\Delta U if q=50Jq=-50\,\text{J} and w=+20Jw=+20\,\text{J} for the system.

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

Flashcard 7: Find qpq_p if n=2n=2 and Cp=30J\cdotpmol1\cdotpK1C_p=30\,\text{J·mol}^{-1}\text{·K}^{-1} for ΔT=10K\Delta T=10\,\text{K}.

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

Flashcard 8: What is the sign convention for ww when the system does work on the surroundings?

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

Flashcard 9: Which sign of ΔH\Delta H indicates an exothermic reaction at constant pressure?

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

Flashcard 10: What is the relationship between ΔG\Delta G^\circ and the equilibrium constant KK?

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

Flashcard 11: What is the sign convention for qq when heat enters the system?

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

Flashcard 12: What is the relationship between CpC_p and CvC_v for an ideal gas?

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

Flashcard 13: State the Gibbs free energy equation relating ΔG\Delta G, ΔH\Delta H, TT, and ΔS\Delta S.

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

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

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

Flashcard 15: What is the relationship between ΔU\Delta U and qq at constant volume?

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

Flashcard 16: Find ΔG\Delta G at T=300KT=300\,\text{K} if ΔH=20kJ\Delta H=20\,\text{kJ} and ΔS=50J\cdotpK1\Delta S=50\,\text{J·K}^{-1}.

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

Flashcard 17: State the ideal gas law relating PP, VV, nn, and TT.

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

Flashcard 18: State the first law of thermodynamics using ΔU\Delta U, qq, and ww.

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

Flashcard 19: Find ww if Pext=2atmP_{\text{ext}}=2\,\text{atm} and the system expands by ΔV=3L\Delta V=3\,\text{L}.

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

Flashcard 20: What is the relationship between ΔH\Delta H and qq at constant pressure (PV work only)?

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

Flashcard 21: What is the criterion for spontaneity in terms of the universe entropy change?

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

Flashcard 22: What is the formula for pressure–volume work at constant external pressure?

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

Flashcard 23: For an isochoric process, what is the work ww?

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

Flashcard 24: What is the formula for heat absorbed at constant volume for a temperature change ΔT\Delta T?

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