Applied Thermodynamics MCQ

Explanation: The question asks for the work performed by one mole of an ideal gas when it expands isothermally from a high to low pressure at a constant temperature of 300 K. Understanding the nature of isothermal processes is key.

In an isothermal process, the temperature remains constant, meaning the internal energy of the gas does not change. For ideal gases, internal energy depends solely on temperature. work done by the gas is equal to the area under the pressure–volume (P–V) curve. The formula for work in an isothermal expansion is W = nRT ln(V_f/V_i), where n is the number of moles, R is the universal gas constant, and V_f and V_i are final and initial volumes, respectively. By using the ideal gas law (PV = nRT), volumes can be expressed in terms of pressures, giving W = nRT ln(P_i/P_f).

This equation shows that the work depends on the temperature, number of moles, and the ratio of initial and final pressures. Conceptually, the gas pushes against the external pressure to expand, converting thermal energy into mechanical work.

For instance, consider one mole of gas in a piston: as it expands slowly, the gas molecules exert force on the piston walls, causing it to move. The total energy spent in pushing the piston corresponds to the work done.

Overall, calculating work in isothermal processes involves understanding the logarithmic relationship between pressure and volume while temperature remains constant.

Explanation: This question asks to identify which statement about thermodynamic processes does not conform to the established principles of Heat, work, and internal energy relationships.

In Thermodynamics, different processes have specific rules: in an isothermal process, the internal energy change ΔU = 0, so Heat added equals work done (q = W). In a cyclic process, the system returns to its initial state, making NET change in internal energy zero, but work and Heat may not directly equate. In an isochoric process (constant volume), no work is done, so the change in internal energy equals the Heat added (ΔU = q). In an adiabatic process, no Heat is exchanged (q = 0), so any change in internal energy equals the negative of work done (ΔU = −W).

Step-by-step, you examine each process: check the definitions of isothermal, adiabatic, cyclic, and isochoric processes. Compare the Heat and work relationships stated in the options with the standard thermodynamic equations. Any statement that violates these fundamental relationships can be identified as incorrect.

For analogy, think of each process as a unique “rulebook” for energy flow. Any statement that contradicts the rulebook is the one that doesn’t belong.

Overall, this question tests conceptual clarity about energy transfer rules in thermodynamic processes.

Explanation: This question asks why a gas cools when it expands adiabatically, meaning no Heat is exchanged with the surroundings.

In an adiabatic process, q = 0, so the system cannot gain or lose Heat. The internal energy of the gas decreases when it performs work on the surroundings to expand. Since temperature is directly related to internal energy for an ideal gas (U = (3/2)nRT for a monatomic gas), a decrease in internal energy causes a drop in temperature.

Step-by-step, as the gas expands, it pushes against external pressure, using its own internal energy to do work. With no Heat entering the system, this energy loss translates directly into reduced kinetic energy of the molecules, hence cooling occurs. The effect is more noticeable in rapid expansions where no thermal equilibrium with the surroundings is established.

For example, when a compressed air cylinder is opened quickly, the escaping air feels cold because adiabatic expansion lowers its temperature.

Overall, the cooling during adiabatic expansion occurs because the energy of the gas is used to do work, decreasing its internal energy and hence its temperature.

Explanation: This question asks how to classify a living organism in terms of thermodynamic systems.

A system can be open, closed, or isolated. An open system exchanges both energy and Matter with its surroundings. A closed system exchanges only energy, not Matter. An isolated system exchanges neither. A homogeneous system has uniform composition but this does not define life in thermodynamic terms.

Step-by-step, Living Organisms take in nutrients and release wastes, exchange gases, and transfer energy (Heat). These interactions with the Environment clearly show exchange of both Matter and energy, fitting the definition of an open system. Closed or isolated systems do not account for Matter exchange, which is essential for metabolism and survival.

An analogy is a boiling pot of soup with a lid partially open: it releases steam and absorbs heat, similar to how Organisms interact with their surroundings.

Overall, Living Organisms function as open systems due to continuous exchange of energy and Matter with their Environment.

Explanation: This question asks to rank gases by work done during identical isothermal expansion, focusing on Molecular Mass and amount of substance.

Work done in isothermal expansion, W = nRT ln(V_f/V_i), depends on the number of moles (n), temperature (T), and volume ratio. For equal masses, gases with lower molar mass have more moles. More moles mean more work done at the same temperature and volume change.

Step-by-step, calculate the moles for 10 g each gas: n = mass / molar mass. CH₄ has the lowest molar mass, so most moles; Cl₂ has the highest molar mass, so fewest moles. Therefore, the work done decreases with decreasing number of moles: CH₄ > O₂ > N₂ > Cl₂.

Analogy: Think of moles as “workers” pushing a piston. More workers (moles) can perform more work.

Overall, for equal Mass gases, work done in isothermal expansion is inversely related to molar Mass due to differences in moles present.

Explanation: The question asks to find work performed by the system using thermodynamic relationships.

The first law of Thermodynamics states ΔU = q − W, where ΔU is change in internal energy, q is heat added to the system, and W is work done by the system. This relationship allows calculation of work done if ΔU and q are known.

Step-by-step, plug values into the formula: ΔU = q − W. Rearrange to find W = q − ΔU. Heat absorbed increases internal energy and/or does work. Subtracting internal energy change from heat absorbed gives work performed by the system.

Example: If a piston absorbs 920 J of heat and internal energy rises by 460 J, the extra energy is used to push the piston outward, performing work.

Overall, the work done corresponds to the portion of absorbed heat not stored as internal energy.

Explanation: This question asks to identify the process that leads to a drop in temperature for a thermodynamic system.

Temperature is related to internal energy. In adiabatic processes, no heat enters or leaves the system. Expansion or compression affects internal energy. Adiabatic expansion reduces internal energy since the system does work on surroundings without heat input, leading to temperature decrease. Adiabatic compression increases internal energy and temperature. Isothermal processes maintain constant temperature.

Step-by-step, analyze options: only adiabatic expansion reduces internal energy without heat compensation. This leads to cooling. Other processes either increase or maintain temperature.

Analogy: Rapidly inflating a bicycle tire with air feels cool because air expands adiabatically, lowering its temperature.

Overall, temperature decreases when a system performs work in adiabatic expansion, reducing internal energy.

Explanation: The question focuses on conditions for equal work done by gases under isothermal expansion with equal masses.

Work done in isothermal expansion depends on the number of moles, temperature, and volume ratio: W = nRT ln(V_f/V_i). Equal masses do not necessarily mean equal moles; moles depend on molar Mass. For equal work, either moles must be equal or conditions adjusted so that W is the same.

Step-by-step, consider pairs of gases: select gases with appropriate molar masses so that equal masses correspond to equal moles. This ensures the same work is performed under identical temperature and pressure conditions.

Example: For equal masses, molecules with higher molar Mass have fewer moles, so to get same work, a lighter gas must balance this.

Overall, equal work is achieved when equal masses correspond to equal effective moles contributing to isothermal expansion.

Explanation: This question asks about energy levels of reactants and products in an exothermic reaction.

Exothermic reactions release energy to surroundings. The energy released is due to the reactants having higher internal energy than the products. The excess energy appears as heat, Light, or work.

Step-by-step, in chemical terms, bond energies of reactants are higher than products; breaking and forming bonds leads to NET release of energy. Products are more stable because they have lower energy than reactants.

Analogy: Think of a ball rolling down a hill—the ball (reactants) has more potential energy initially and loses it while rolling to a lower position (products), releasing energy to the Environment.

Overall, in exothermic reactions, reactants are higher in energy compared to products, which allows energy release during the reaction.

Explanation: The question asks to classify the thermal nature of the evaporation process.

Evaporation requires molecules at the surface to overcome intermolecular forces and escape into the gas phase. Energy is absorbed from the surroundings to facilitate this phase change, making it an endothermic process. Temperature of the remaining liquid may decrease due to energy loss.

Step-by-step, energy input from surroundings increases the kinetic energy of molecules, enabling escape. No chemical reaction occurs; only physical change. Heat is absorbed continuously until equilibrium vapor pressure is reached.

Analogy: Sweat evaporation cools the skin because energy is absorbed from the body, similar to the liquid absorbing heat from surroundings.

Overall, evaporation is an energy-absorbing (endothermic) physical process involving phase change from liquid to vapor.

Explanation: The question asks for work performed by a gas expanding into a vacuum, known as free expansion.

In free expansion, the gas expands without resistance; the external pressure is zero. Work done is calculated as W = ∫ P_ext dV. Since P_ext = 0, the integral equals zero. Internal energy does not change for ideal gases in free expansion because temperature remains constant.

Step-by-step, the gas molecules spread out to occupy the available volume, but no energy is transferred to push against surroundings. This differentiates free expansion from expansion against a finite pressure, where work is done.

Example: Opening a piston into an evacuated chamber allows gas to fill it instantly without performing work.

Overall, ideal gas expansion into vacuum involves zero work, despite change in volume.