An ideal diatomic gas is operated as volume*(temperature)^3=constant. If final temperature is 3 times the initial temperature find change in internal energy and work done by gas

To tackle the problem of an ideal diatomic gas where the relationship between volume and temperature is given by the equation \( V \cdot T^3 = \text{constant} \), we need to analyze the situation step by step. We are asked to find the change in internal energy and the work done by the gas when the final temperature is three times the initial temperature.

Understanding the Relationship

First, let's denote the initial temperature as \( T_1 \) and the final temperature as \( T_2 = 3T_1 \). According to the relationship provided, we can express the volumes at these temperatures as follows:

• Initial volume: \( V_1 \cdot T_1^3 = C \) (where \( C \) is a constant)
• Final volume: \( V_2 \cdot T_2^3 = C \)

From these equations, we can derive the volumes:

• From the initial condition: \( V_1 = \frac{C}{T_1^3} \)
• From the final condition: \( V_2 = \frac{C}{(3T_1)^3} = \frac{C}{27T_1^3} \)

Calculating the Change in Internal Energy

The change in internal energy (\( \Delta U \)) for an ideal gas can be calculated using the formula:

ΔU = nC_vΔT

For a diatomic gas, the molar heat capacity at constant volume (\( C_v \)) is typically \( \frac{5}{2}R \). The change in temperature (\( ΔT \)) is:

ΔT = T_2 - T_1 = 3T_1 - T_1 = 2T_1

Now substituting this into the equation for \( ΔU \):

ΔU = n \cdot \frac{5}{2}R \cdot (2T_1) = 5nRT_1

Determining the Work Done by the Gas

Next, we need to find the work done by the gas during this process. The work done by the gas in a quasi-static process can be expressed as:

W = ∫ P dV

However, since we have a specific relationship between volume and temperature, we can use the ideal gas law \( PV = nRT \) to express pressure in terms of volume and temperature:

P = \frac{nRT}{V}

To find the work done, we can substitute this expression into the work integral. But first, we need to find the change in volume:

ΔV = V_2 - V_1 = \frac{C}{27T_1^3} - \frac{C}{T_1^3} = C\left(\frac{1}{27T_1^3} - \frac{1}{T_1^3}\right) = C\left(\frac{1 - 27}{27T_1^3}\right) = -\frac{26C}{27T_1^3}

Now, the work done can be approximated using the average pressure and the change in volume:

W = P_{avg} \cdot ΔV

Using the ideal gas law at the initial state, we can find the average pressure:

P_{avg} = \frac{nR(T_1 + T_2)}{2V_{avg}} = \frac{nR(4T_1)}{2\left(\frac{C}{T_1^3}\right)} = \frac{2nRT_1^4}{C}

Substituting this into the work equation gives:

W = \frac{2nRT_1^4}{C} \cdot \left(-\frac{26C}{27T_1^3}\right) = -\frac{52nRT_1}{27}

Final Results

To summarize, we have:

• Change in internal energy (ΔU): \( 5nRT_1 \)
• Work done by the gas (W): \( -\frac{52nRT_1}{27} \)

This analysis shows how the relationships between volume, temperature, and the properties of an ideal gas can be used to derive important thermodynamic quantities. If you have any further questions or need clarification on any part of this process, feel free to ask!