Problem 175 - A Complete Resource Book in Chemistry for JEE Main

Question

A piston filled with \(0.04\) mol of an ideal gas expands reversibly from \(50.0 \mathrm{~mL}\) to \(375 \mathrm{~mL}\) at a constant temperature of \(37.0^{\circ} \mathrm{C}\). As it does so, it absorbs \(208 \mathrm{~J}\) of heat. The values of \(\mathrm{q}\) and \(\mathrm{w}\) for the process will be: \((\mathrm{R}=3.314 \mathrm{~J} / \mathrm{mol} \mathrm{K})(\operatorname{Ln} 7.5=2.01)\) (a) \(\mathrm{q}=-208 \mathrm{~J}, \mathrm{w}=+208 \mathrm{~J}\) (b) \(\mathrm{q}=+208 \mathrm{~J}, \mathrm{w}=+208 \mathrm{~J}\) (c) \(\mathrm{q}=+208 \mathrm{~J}, \mathrm{w}=-208 \mathrm{~J}\) (d) \(\mathrm{q}=-208 \mathrm{~J}, \mathrm{w}=-208 \mathrm{~J}\)

Step-by-Step Solution

Answer

(c) \(q=+208\, \text{J}, w=-208\, \text{J}\).

Step 1: Identify the type of process

The problem states an ideal gas undergoing a reversible expansion at constant temperature (isothermal). This means the process is isothermal.

Step 2: Analyze the sign of heat (q)

The problem states that the system absorbs 208 J of heat. In thermodynamics, heat absorbed by the system is positive: \(q = +208\, \text{J}\).

Step 3: Apply the First Law of Thermodynamics

The First Law of Thermodynamics is \( \Delta U = q + w \). In an isothermal process for an ideal gas, the change in internal energy \(\Delta U\) is zero because temperature is constant.

Step 4: Calculate work (w) using \( \Delta U = 0 \)

Since \( \Delta U = 0 \) for an isothermal process, we have \(0 = q + w\). Thus, \(w = -q\). Substituting for \(q\): \(w = -208 \, \text{J}\).

Step 5: Verify the math

Recheck that \(0 = q + w\) holds true. Substituting the values, \(0 = 208 + (-208)\), thus ensuring both \(q\) and \(w\) adhere to the First Law.

Key Concepts

Isothermal Process

In an isothermal process, the temperature remains constant throughout the operation of the system. This is a key feature affecting the behavior and properties of the system. For an ideal gas undergoing an isothermal process, the relationship between pressure and volume can be described by Boyle's Law, where - As the volume of the gas increases, the pressure decreases, provided the temperature stays unchanged. The temperature constancy also means there's no change in internal energy (denoted by \( \Delta U \)), for an ideal gas. Therefore, any heat absorbed by the system would result in work done by the system to the surroundings or vice versa. During such a process, it is vital to understand that: - Isothermal (constant temperature) processes require slow and careful adjustments to ensure uniform temperature across the gas. - The internal energy remains constant, so any heat flow into the system manifests as work done by the system.

Ideal Gas Law

The ideal gas law combines several gas laws into one comprehensive equation: \( PV = nRT \). This law provides a simplified model to predict the behavior of gases. In the equation: - \( P \) stands for pressure, - \( V \) is volume, - \( n \) represents the number of moles, - \( R \) is the ideal gas constant, and - \( T \) is the temperature in Kelvin. In isothermal conditions, where temperature stays constant, the ideal gas law helps in understanding the inverse relationship between pressure and volume (Boyle's Law). Ideal gases ideally fit this model, allowing simplifications such as constant internal energy during isothermal changes. This model is useful for making approximations in many real-world applications, especially when dealing with gases at high temperatures and low pressures.

First Law of Thermodynamics

The First Law of Thermodynamics is a version of the law of conservation of energy. It states that the energy of an isolated system is constant. In formula terms, it is expressed as \[ \Delta U = q + w \] where - \( \Delta U \) is the change in internal energy, - \( q \) is the heat added to the system, and - \( w \) is the work done by the system. For an isothermal process involving an ideal gas, \( \Delta U \) is zero because the temperature remains unchanged. Thus, any energy introduced as heat would precisely be offset by the work done by the system, meaning \( q = -w \). Understanding this relationship helps solve scenarios where energy transactions occur without altering the internal energy, simplifying the calculations and predictions of energy flow.