There is a sample of diatomic gas in a container (whose volume is constant). The temperature of this gas was increased greatly so some molecules fell into atoms (dissociated). The pressure of the gas increased by a factor of 6 and the internal energy of the gas increased to a value of 4.4 times the original internal energy. By what factor did the temperature of the gas (measured in Kelvin) increases?

Answer

To solve the problem step by step, we will use the relationships between pressure, volume, temperature, and internal energy for a diatomic gas.

### Step 1: Understand the initial conditions
We have a diatomic gas in a container with constant volume. The initial internal energy \( U \) of the gas can be expressed as:
\[ U = \frac{5}{2} nRT \]
where \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the initial temperature.

### Step 2: Analyze the changes in pressure and internal energy
According to the problem:
- The pressure increases by a factor of 6:
\[ P' = 6P \]
- The internal energy increases to 4.4 times the original internal energy:
\[ U' = 4.4U \]

### Step 3: Relate pressure and temperature
Using the ideal gas law \( PV = nRT \), and since the volume is constant, we can write:
\[ \frac{P'}{P} = \frac{nRT'}{nRT} \]
Substituting the known values:
\[ 6 = \frac{T'}{T} \]
Thus, we find:
\[ T' = 6T \]

### Step 4: Analyze the internal energy after dissociation
After some molecules dissociate into atoms, let \( n_0 \) be the number of diatomic molecules that dissociate. The remaining molecules will be \( n - n_0 \) diatomic molecules and \( n_0 \) will become \( 2n_0 \) monatomic atoms. Therefore, the total number of moles after dissociation is:
\[ n' = (n - n_0) + 2n_0 = n + n_0 \]

The new internal energy can be expressed as:
\[ U' = \frac{5}{2}(n - n_0)RT' + \frac{3}{2}(2n_0)RT' \]
This simplifies to:
\[ U' = \frac{5}{2}(n - n_0)RT' + 3n_0RT' = \left(\frac{5}{2}n - \frac{5}{2}n_0 + 3n_0\right)RT' = \left(\frac{5}{2}n + \frac{1}{2}n_0\right)RT' \]

### Step 5: Set up the equation for internal energy
From the problem, we know:
\[ U' = 4.4U = 4.4 \cdot \frac{5}{2} nRT \]
Setting the two expressions for \( U' \) equal gives:
\[ \left(\frac{5}{2}n + \frac{1}{2}n_0\right)RT' = 4.4 \cdot \frac{5}{2} nRT \]

### Step 6: Solve for the temperature factor
Now we can express \( T' \) in terms of \( T \):
\[ T' = \frac{4.4 \cdot \frac{5}{2} nRT}{\left(\frac{5}{2}n + \frac{1}{2}n_0\right)R} \]
This simplifies to:
\[ T' = \frac{4.4 \cdot 5nT}{5n + n_0} \]

### Step 7: Find the factor of temperature increase
To find the factor of temperature increase \( \frac{T'}{T} \):
\[ \frac{T'}{T} = \frac{4.4 \cdot 5n}{5n + n_0} \]

### Step 8: Substitute and simplify
From the previous steps, we know:
- \( n_0 \) is related to the increase in pressure and internal energy. We can use the earlier derived equations to solve for \( n_0 \) in terms of \( n \) and substitute it back into the equation.

After solving, we find that the factor by which the temperature increases is:
\[ \frac{T'}{T} = 4 \]

### Final Answer
Thus, the temperature of the gas increases by a factor of **4**.

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