Question

For Diatomic gas, find the ratio of \(\Delta Q : \Delta U : W\) for an isobaric process.

Hint:

In an isobaric process, the heat added to the system is split between changing the internal energy and doing work. For diatomic gases, the ratio of these quantities can be derived from the heat capacities at constant volume and pressure.

Answer 1

Correct Answer: 2

For an isobaric process (a process occurring at constant pressure), the first law of thermodynamics is given by:

\[\Delta Q = \Delta U + W\]

Where: 
- \(\Delta Q\) is the heat added to the system,
- \(\Delta U\) is the change in internal energy,
- \(W\) is the work done by the system.

For a diatomic gas, the work done in an isobaric process is:

\[W = P \Delta V\]

 

Also, the change in internal energy for a diatomic gas (which behaves like an ideal gas) is given by:

\[\Delta U = n C_V \Delta T\]

 

The heat added to the system in an isobaric process is:

\[\Delta Q = n C_P \Delta T\]

 

From thermodynamics, the relationship between \(C_P\) and \(C_V\) for an ideal gas is:

\[C_P = C_V + R\]

 

For a diatomic ideal gas, \(C_V = \frac{5}{2}R\) and \(C_P = \frac{7}{2}R\).

Now, we can calculate the ratio of \(\Delta Q : \Delta U : W\):

\[\Delta Q = n C_P \Delta T = n \left( \frac{7}{2}R \right) \Delta T\]

\[\Delta U = n C_V \Delta T = n \left( \frac{5}{2}R \right) \Delta T\]

\[W = P \Delta V = n R \Delta T\]

Thus, the ratio is:

\[\Delta Q : \Delta U : W = \left( \frac{7}{2} \right) : \left( \frac{5}{2} \right) : 1\]

Simplifying this, we get:

\[\Delta Q : \Delta U : W = 7 : 5 : 2\]

Therefore, the correct ratio is 7:5:2.