Question

If the ratio of specific heat of a gas at Constant pressure to that at constant volume is \(\gamma\), the Change in internal energy of the mass of gas, when the volume changes from \(\mathrm{V}\) to \(2 \mathrm{~V}\) at Constant Pressure p, is (A) \(\\{(\mathrm{PV}) /(\gamma-1)\\}\) (B) \(\\{\mathrm{R} /(\gamma-1)\\}\) (C) PV (D) \(\\{(\gamma \mathrm{PV}) /(\gamma-1)\\}\)

Short answer

The change in internal energy of the mass of gas when the volume changes from V to 2V at constant pressure p is given by (D) \(\frac{\gamma PV}{\gamma-1}\).

Step-by-step solution

Write down the given information

The ratio of specific heat at constant pressure (Cp) to that at constant volume (Cv) is given by \(\gamma = \frac{C_{p}}{C_{v}}\). The volume changes from V to 2V under constant pressure p.

Ideal Gas Law

Via the ideal gas law, we know that \(PV=nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since we have constant pressure, the relationship between the initial and final states is: \[ \frac{P(2V)}{PV}=\frac{T_{2}}{T_{1}} \implies T_{2} = 2T_{1} \]

Specific Heats and Internal Energy relationship

Recall the definition of specific heats at constant volume and constant pressure, \(C_{v}=\frac{dQ}{dT}\), \(C_{p}=\frac{dQ}{dT}\), where dQ is heat transferred, and dT is the change in temperature. Using the information given about the specific heat ratio, which is \(\gamma=\frac{C_{p}}{C_{v}}\), Combine these to obtain: \[ C_{v} =\frac{1}{\gamma-1} C_{p} \] Also, recall the relationship between internal energy (\(dU\)), heat transferred (\(dQ\)), and work done (\(dW\)): \(dU=dQ-dW\). When the process is carried out at constant pressure, work done is given by, \[ dW = PdV \]

Substitute the expressions and solve for internal energy change

Now, the change in internal energy is given by: \[ \Delta U=\int_{T_{1}}^{T_{2}} C_{v}dT \] Substitute the expressions from step 3 for \(C_{v}\) in terms of \(C_{p}\) and step 2 for initial and final temperatures: \[ \Delta U = \frac{1}{\gamma-1}\int_{T_{1}}^{2T_{1}} C_{p} dT \] Since \(C_{p}\) is constant, it can be taken out of the integral: \[ \Delta U = \frac{1}{\gamma-1} C_{p} \int_{T_{1}}^{2T_{1}} dT \] Evaluate the integral: \[ \Delta U=\frac{1}{\gamma-1} C_{p} (2T_{1}-T_{1})= \frac{1}{\gamma-1} C_{p} T_{1} \] Now, substitute the ideal gas law expression for the initial state, \(PV=nRT_{1}\): \[ \Delta U = \frac{1}{\gamma-1}\frac{C_{p}}{R} PV \] Finally, since \(C_{p}=\gamma C_{v}\), and since \(C_{p}-C_{v}=R\), we have \(C_{p}=\gamma R\). Substitute this into the previous equation to obtain the final answer: \[ \Delta U = \frac{\gamma PV}{\gamma-1} \] Thus, the correct answer is (D) \(\frac{\gamma PV}{\gamma-1}\).

Key concepts

Specific Heat Ratio

The specific heat ratio, denoted by \( \gamma \), is an important parameter in thermodynamics, particularly for gases. This ratio is defined as the specific heat capacity at constant pressure \( (C_p) \) divided by the specific heat capacity at constant volume \( (C_v) \). Mathematically, it is expressed as:

• \( \gamma = \frac{C_p}{C_v} \)

This ratio is crucial when analyzing processes involving gases, because it provides insights into how much heat energy is required to raise the temperature of a gas under different constraints.
A higher \( \gamma \) indicates that a gas requires more energy to increase its temperature when held at constant pressure compared to when it is held at constant volume. This property affects the behavior and efficiency of engines and other systems where gases undergo thermal processes.

Ideal Gas Law

The ideal gas law is a fundamental principle that describes the relationship between pressure, volume, and temperature for an ideal gas. It is commonly written as:

• \( PV = nRT \)

Here:

• \( P \) is the pressure
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the universal gas constant
• \( T \) is the temperature in Kelvin

This equation helps predict how a gas will behave under varying conditions. For example, if a gas is subjected to constant pressure and its volume doubles, the ideal gas law can be used to determine the change in temperature.
In the context of the exercise, this law plays a key role in understanding how internal energy changes as the volume of a gas transitions from \( V \) to \( 2V \) while maintaining constant pressure.

Work Done at Constant Pressure

When a gas expands or contracts at constant pressure, work done by or on the gas can be calculated using the formula:

• \( dW = PdV \)

Here:

• \( dW \) is the work done on the gas
• \( P \) is the constant pressure
• \( dV \) is the change in volume

This expression reflects that the work done depends on the pressure and the change in volume of the gas.
Understanding this concept is essential for solving problems related to thermodynamic processes. During a constant pressure process, when a gas expands, it does positive work on its surroundings, losing internal energy depending on how much volume changes.

Relationship Between Internal Energy and Work

The internal energy of a gas is the total energy contained within its molecules and can change due to heat transfer and work done. The First Law of Thermodynamics expresses this relationship as:

• \( dU = dQ - dW \)

Where:

• \( dU \) is the change in internal energy
• \( dQ \) is the heat added to the system
• \( dW \) is the work done by the system

This relationship shows that internal energy increases with added heat and decreases when work is done by the system. When considering gases, it is important to note that specific heat capacities \( (C_v \) and \( C_p) \), relate internal energy changes to temperature changes.
In the constant pressure scenario as mentioned in the exercise, internal energy change can be derived by combining the specific heat capacities and the work done expressions to understand overall energy changes efficiently.