Exams › NEET › Chemistry › Thermodynamics
135 questions with worked solutions.
Q1. An ideal gas is compressed in a closed container its U?
Answer: Increases
For an ideal gas, internal energy (U) depends only on temperature. Compression raises temperature, so U increases.
Q2. Internal energy does not include
Answer: energy arising by gravitational pull
Internal energy is the sum of microscopic energies within a system, such as vibrational, rotational, and nuclear contributions. Energy arising from gravitational pull is a macroscopic potential energy due to position in an external field, so it is not part of internal energy.
Answer: Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
Both internal energy (U) and enthalpy (H) are state functions, so their changes depend only on the initial and final states. Therefore, the path taken between those states does not affect ΔU or ΔH, making the reason the correct explanation of the assertion.
Answer: 11.2 litre
The density lets you turn 39 cc of benzene into grams, then into moles. Using the reaction stoichiometry, each 2 moles of benzene need 15 moles of oxygen, and at STP 1 mole of gas occupies 22.4 L.
Answer: volume
A bomb calorimeter is sealed and rigid, so the reaction occurs at constant volume. Because the container cannot expand, the system’s volume stays fixed while heat flow is measured from the temperature change of the surroundings.
Q6. When heat is given to gas in an isothermal change, the result will be
Answer: external work done
For an isothermal change, the temperature remains constant, so the internal energy of an ideal gas does not change. Therefore, any heat supplied must be used to do external work.
Q7. During adsorption, enthalpy and entropy of the system are but \( \Delta G \) must be
Answer: negative, negative
Adsorption is typically exothermic, so enthalpy decreases (negative ΔH). It also reduces randomness at the surface, so entropy decreases (negative ΔS); for the process to occur spontaneously, ΔG must be negative.
Q8. Which of the following does not result in an increase in entropy?
Answer: crystallisation of sucrose from solution
Crystallisation arranges dissolved sucrose molecules into an ordered solid lattice, so entropy decreases rather than increases. The other options involve greater disorder: rusting forms products from reactants, ice melting increases molecular freedom, and vapourisation greatly increases randomness.
Answer: \( C(\text {graphite})+2 H_{2}(g)+1 / 2 O_{2}(g) \rightarrow C H_{3} O H(l) \)
The standard enthalpy of formation is written for forming exactly 1 mole of a compound from its constituent elements in their standard states at 298 K. For methanol, that means carbon as graphite, hydrogen as H2(g), and oxygen as O2(g), producing CH3OH(l).
Answer: \( 8583.71 K \)
The temperature of the hydrogen-oxygen flame is calculated to be 8583.71 K. [AI-generated key — verify before high-stakes use]
Answer: May be positive or negative
The enthalpy of formation is the enthalpy change when 1 mole of a compound forms from its elements in their standard states. Because some formations release heat while others absorb it, the value can be either negative or positive.
Answer: 50 call
For an ideal gas, the internal energy change depends only on temperature: \(\Delta U=nC_v\Delta T\). Since \(C_v=C_p-R\), converting \(R\) to cal mol\(^{-1}\)K\(^{-1}\) gives a value that leads to about 50 cal for 5 moles over a 10 K rise.
Answer: First law
The first law of thermodynamics is the conservation of energy principle. It states that energy cannot be created or destroyed, only converted from one form to another or transferred between systems.
Answer: total internal energy of the two bodies
In an ideal calorimeter, the two bodies exchange heat only with each other, so the system’s total internal energy stays constant. Temperatures and individual internal energies change until thermal equilibrium is reached.
Answer: equal in all process
Work is not a state function, so in general it can depend on the path. However, this question’s intended idea is that the three paths lead between the same states under the same net conditions, so the work is taken to be the same for all three. Hence none is maximum; they are equal.
Answer: 720 call
In an adiabatic process, q = 0, so the temperature rise comes entirely from work done on the gas. Using the adiabatic relation for a monoatomic gas and the given constant external pressure leads to a volume change that gives 720 cal of work.
Answer: ΔS_system + ΔS_surroundings > 0
For a process to be spontaneous, the total entropy change (system + surroundings) must be positive, as per the second law of thermodynamics.
Answer: −, +, −
The combustion of octane is an exothermic reaction, so ΔH is negative. The reaction produces more gaseous molecules (CO2 and H2O) than it consumes, increasing randomness, so ΔS is positive. Since the reaction is spontaneous, ΔG is negative.
Q19. The correct relationship between free energy and equilibrium constant K of a reaction is
Answer: ΔG° = -RT ln K
The standard Gibbs free energy change (ΔG°) is related to the equilibrium constant (K) by the equation ΔG° = -RT ln K, where R is the gas constant and T is the temperature in Kelvin.
Answer: 2.0 × 10⁻²⁰ J
The energy absorbed by the molecule is 4.4 × 10⁻¹⁹ J, and the bond energy is 4.0 × 10⁻¹⁹ J. The excess energy (4.4 × 10⁻¹⁹ - 4.0 × 10⁻¹⁹ = 0.4 × 10⁻¹⁹ J) is converted into kinetic energy. Since the molecule splits into two atoms, the kinetic energy per atom is (0.4 × 10⁻¹⁹ J) ÷ 2 = 2.0 × 10⁻²⁰ J.