NEET Physics: work done on compressing a gas Practice Questions

Correct answer: C — Pressure and volume.

Work done in a volume change is fundamentally defined by the change in pressure and volume (W = ∫ P dV). Compression involves decreasing volume, which results in a corresponding change in pressure and volume over that path.

Correct answer: D — First Law of Thermodynamics

The First Law of Thermodynamics states that the change in internal energy of a system is equal to the heat supplied to the system minus the work done by the system (ΔU = Q - W). If work is done *on* the gas, W is positive when considering the system's change, or ΔU = Q + W depending on the sign convention; the fundamental relationship is the First Law.

Correct answer: D — nRT ln(V₂/V₁)

For a reversible process (which is the basis for the simplest conceptual work calculation in KTG), the work done on the gas during compression is W = nRT ln(V₂/V₁). This formula relates the change in internal energy to the pressure-volume change when considering the change in kinetic energy of the molecules.

Correct answer: D — Positive

Work done on the gas during compression (W = -PΔV) is negative if we consider the standard convention where work done *by* the gas is positive. However, if we consider the work done *on* the gas by an external agent (compression), the work done *on* the system is positive, as the internal energy increases (assuming constant temperature is not explicitly stated, focusing purely on volume change). For compression, ΔV is negative, and Wₒₙ = -PΔV = P|ΔV|, which is positive work done *on* the gas.

Correct answer: C — Total internal energy

The work done on the gas during compression directly changes the system's internal energy (ΔU = W).

Correct answer: C — The relationship between work and internal energy.

The work done on the gas directly changes its internal energy (ΔU). For an ideal gas, ΔU is related to the change in kinetic energy, which is fundamentally tied to the macroscopic work done during the process.

Correct answer: D — Work done on the gas during compression is less than the work done during free expansion.

Isothermal compression involves external work being done against the gas pressure, leading to a positive work term W = -nRT ln(V_f/Vᵢ). Free expansion is an expansion against zero external pressure, meaning W=0. Therefore, the work done during compression is positive (work done *on* the gas), while the work done during free expansion is zero.

Correct answer: B — The work done is equal to the negative change in internal energy.

For a reversible process, the change in internal energy (ΔU) is related to the heat supplied (Q) and the work done (W) by the First Law of Thermodynamics: ΔU = Q - W. If we consider work done *on* the gas, W is negative, thus ΔU = Q + |W|. If we focus purely on the work component related to internal energy change, the work done on the system is W. If the compression is adiabatic (Q=0), then ΔU = -W (where W is work done *by* the gas). In the context of work done *on* the gas, ΔU = -W (if W is defined as work done by the system). Option 2 is the most general form of the First Law.

Correct answer: B — Constant temperature

Adiabatic processes are characterized by no heat exchange (Q=0). Since the compression is irreversible (or nearly so), the process is defined by the relationship between pressure, volume, and temperature changes, which are complex, but the process itself is defined by the lack of heat transfer.

Correct answer: B — The work done is proportional to the product of pressure and the change in volume.

Work done (W) during a compression of an ideal gas is W = ∫ P dV. Since pressure (P) is positive and volume change (dV) is negative during compression, the work done *on* the gas is positive. Higher pressure leads to a larger work done for the same volume change.

Correct answer: B — Always greater than the reversible work.

For any real (irreversible) process, the work done is always greater than the work done in the corresponding reversible process connecting the same initial and final states.

Correct answer: B — Negative value

During compression, work is done *on* the system, so the work done on the gas is positive. However, the work done *by* the gas is negative. In physics convention, work done *on* the gas is positive, but the change in internal energy is related to this work.

Correct answer: D — P₁(V₂ - V₁)

Work done (W) by an external agent on a gas during a volume change is defined by the integral of pressure with respect to volume. For a process involving pressure changes, the work done is W = ∫ P dV. If the process is simple compression (ignoring the path details for the basic concept), the work done is related to the change in volume multiplied by the appropriate pressure (depending on the context). In a general context where pressure changes, the work is related to the pressure change over the volume change. However, for the simplest conceptual recall of work done, we relate it to the pressure change applied over the volume change. The work done *on* the gas during compression is fundamentally ∫ P dV. For an easy conceptual check, if we consider the work done based on the pressure difference over the displacement (as often simplified in introductory problems): W = P₍avg) ΔV. If we assume P₁ is the initial pressure and V₂ - V₁ is the volume change, the simplest conceptual relationship often tested is W = P(V₂ - V₁) when pressure is assumed constant during the simple displacement, but the general thermodynamic definition requires integration. For this level, the work done is directly related to the change in the system's state. In the context of simple work done by compression, the change in energy is related to pressure times the change in volume.

Correct answer: B — Internal Energy

For a reversible adiabatic process (adiabatic compression), the work done on the gas is equal to the negative change in internal energy (W = -ΔU). Since the process is adiabatic (no heat exchange), the relationship involves the temperature change, and the work is fundamentally related to the change in the state of the gas, which is quantified by its internal energy change.

Correct answer: C — The change in internal energy is equal to the work done done on the gas.

According to the First Law of Thermodynamics (ΔU = Q - W), for a reversible process involving an ideal gas, the work done (W) is directly related to the change in internal energy (ΔU) and the heat exchanged. In many basic contexts, the work done on the gas directly changes its internal energy (assuming Q=0 for adiabatic compression, or relating W to ΔU).