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GATE Life Sciences Thermodynamics Flashcards
50 question-and-answer cards covering Thermodynamics as it is examined in GATE Life Sciences. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Thermodynamics deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
Define the standard enthalpy (heat) of formation, $\Delta H_f^{\circ}$.
The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states (most stable form) at $298\ \mathrm{K}$ and $1\ \mathrm{bar}$.
What is the standard enthalpy of formation of an element in its most stable form?
It is zero by convention, e.g. $\Delta H_f^{\circ}(\ce{O2}, g) = 0$, $\Delta H_f^{\circ}(\ce{C}, graphite) = 0$.
Express the standard enthalpy of reaction in terms of enthalpies of formation.
$$\Delta H_{rxn}^{\circ} = \sum n\,\Delta H_f^{\circ}(\text{products}) - \sum n\,\Delta H_f^{\circ}(\text{reactants})$$ where $n$ are stoichiometric coefficients.
Define the standard enthalpy of combustion.
The enthalpy change when one mole of a substance is completely burned (oxidised) in excess oxygen under standard conditions. It is always negative (combustion is exothermic).
Define entropy ($S$) qualitatively.
Entropy is a state function that measures the degree of randomness or disorder of a system, or equivalently the number of accessible microstates. Higher disorder corresponds to higher entropy.
Give the thermodynamic definition of an infinitesimal entropy change for a reversible process.
$$dS = \frac{dq_{rev}}{T}$$ Entropy change equals the heat absorbed reversibly divided by the absolute temperature.
Write the Boltzmann (statistical) definition of entropy.
$$S = k_{B}\ln W$$ where $k_{B}$ is the Boltzmann constant and $W$ is the number of microstates (thermodynamic probability) corresponding to the macrostate.
State the second law of thermodynamics in terms of entropy.
For any spontaneous (irreversible) process the total entropy of the universe increases: $\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} > 0$. For a reversible process it stays constant.
State the third law of thermodynamics.
The entropy of a perfect, pure crystalline substance is zero at absolute zero temperature ($T = 0\ \mathrm{K}$), i.e. $\lim_{T \to 0} S = 0$. This allows absolute entropies to be determined.
Write the entropy change for an isothermal reversible expansion of an ideal gas.
$$\Delta S = nR\ln\frac{V_{2}}{V_{1}} = nR\ln\frac{P_{1}}{P_{2}}$$ Entropy increases on expansion.
Write the formula for entropy change of a phase transition at its transition temperature.
$$\Delta S = \frac{\Delta H_{trans}}{T_{trans}}$$ e.g. for fusion $\Delta S_{fus} = \Delta H_{fus}/T_{m}$ and for vaporization $\Delta S_{vap} = \Delta H_{vap}/T_{b}$.
How does entropy generally change for the physical states solid, liquid and gas?
Entropy increases in the order $S_{solid} < S_{liquid} < S_{gas}$, because disorder and the number of accessible microstates increase from solid to gas.
Predict the sign of $\Delta S$ for $\ce{2H2(g) + O2(g) -> 2H2O(l)}$.
$\Delta S < 0$ (negative). Gaseous reactants (3 mol of gas) become a liquid product with fewer moles and far less disorder.
Define Gibbs free energy ($G$) and write its defining equation.
Gibbs free energy is a state function representing the maximum non-expansion (useful) work obtainable at constant $T$ and $P$, defined as $$G = H - TS.$$
Write the equation for Gibbs free energy change at constant temperature.
$$\Delta G = \Delta H - T\Delta S$$ This is the Gibbs–Helmholtz relation used to judge spontaneity at constant $T$ and $P$.
What is the spontaneity criterion using Gibbs free energy at constant $T$ and $P$?
If $\Delta G < 0$ the process is spontaneous; if $\Delta G = 0$ the system is at equilibrium; if $\Delta G > 0$ the process is non-spontaneous (reverse is spontaneous).
Define Helmholtz free energy ($A$ or $F$).
$$A = U - TS$$ It is the maximum total work obtainable at constant temperature and volume; $\Delta A < 0$ indicates spontaneity at constant $T$ and $V$.
Make a table of $\Delta H$, $\Delta S$ signs and when a reaction is spontaneous via $\Delta G = \Delta H - T\Delta S$.
$\Delta H<0,\ \Delta S>0$: spontaneous at all $T$. $\Delta H>0,\ \Delta S<0$: non-spontaneous at all $T$. $\Delta H<0,\ \Delta S<0$: spontaneous at low $T$. $\Delta H>0,\ \Delta S>0$: spontaneous at high $T$.
At what temperature does a reaction switch between spontaneous and non-spontaneous?
At the temperature where $\Delta G = 0$, i.e. $$T = \frac{\Delta H}{\Delta S}$$ (the equilibrium/crossover temperature).
Relate standard Gibbs free energy change to the equilibrium constant.
$$\Delta G^{\circ} = -RT\ln K$$ A large positive $K$ corresponds to a large negative $\Delta G^{\circ}$.
Relate standard Gibbs free energy change to standard cell potential (electrochemistry).
$$\Delta G^{\circ} = -nFE^{\circ}_{cell}$$ where $n$ is moles of electrons transferred and $F$ is the Faraday constant. A positive $E^{\circ}_{cell}$ gives a spontaneous cell reaction.
Write the relation between $\Delta G$ under non-standard conditions and the reaction quotient $Q$.
$$\Delta G = \Delta G^{\circ} + RT\ln Q$$ At equilibrium $Q = K$ and $\Delta G = 0$, recovering $\Delta G^{\circ} = -RT\ln K$.
For an ideal gas free (Joule) expansion into vacuum, what are $w$, $q$, and $\Delta U$?
Free expansion is against zero external pressure, so $w = 0$. For an ideal gas $\Delta U = 0$ (isothermal), therefore $q = 0$. Although $q=0$, the process is irreversible and $\Delta S_{system} = nR\ln(V_2/V_1) > 0$.
Why must $\Delta H - T\Delta S$ be negative rather than just $\Delta H$ for predicting spontaneity?
Because spontaneity depends on the total entropy change of the universe. $\Delta G = \Delta H - T\Delta S$ combines the system's enthalpy (related to surroundings' entropy via $-\Delta H/T$) and the system's entropy into one criterion; $-\Delta G/T = \Delta S_{universe}$, so $\Delta G<0 \iff \Delta S_{universe}>0$.
What this deck covers
The Thermodynamics deck follows the GATE Life Sciences Thermodynamics syllabus — 6 chapters and 9 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 8.3 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 162 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Thermodynamics flashcards FAQ
How many Thermodynamics flashcards are in this GATE Life Sciences deck?
50 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these GATE Life Sciences flashcards free?
Yes. The preview here is free to read with no signup, and the full 50-card deck is free inside the Examius app.
What do the Thermodynamics cards cover?
They follow the GATE Life Sciences Thermodynamics syllabus — 6 chapters and 9 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.