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, we need to analyze the changes in pressure, internal energy, and temperature of the diatomic gas during the dissociation process. Let's break it down step by step.

### Step 1: Understand the Initial Conditions
- Let the initial internal energy of the diatomic gas be \( U_0 \).
- The initial temperature of the gas is \( T_0 \).
- The initial number of moles of diatomic gas is \( n \).

### Step 2: Analyze the Changes
- The pressure of the gas increases by a factor of 6:
\[ P_f = 6P_0 \]
- The internal energy increases to 4.4 times the original internal energy:
\[ U_f = 4.4 U_0 \]

### Step 3: Relate Internal Energy to Temperature
For a diatomic gas, the internal energy is given by:
\[ U = \frac{5}{2} nRT \]
Thus, the initial internal energy can be expressed as:
\[ U_0 = \frac{5}{2} nRT_0 \]
The final internal energy can be expressed as:
\[ U_f = \frac{5}{2} nRT_f \]

### Step 4: Set Up the Internal Energy Equation
From the information given, we have:
\[ U_f = 4.4 U_0 \]
Substituting the expressions for \( U_f \) and \( U_0 \):
\[ \frac{5}{2} nRT_f = 4.4 \left( \frac{5}{2} nRT_0 \right) \]
Cancelling \( \frac{5}{2} nR \) from both sides gives:
\[ T_f = 4.4 T_0 \]

### Step 5: Relate Pressure to Temperature
Using the ideal gas law, we know:
\[ PV = nRT \]
Since the volume is constant, we can relate the initial and final states:
\[ \frac{P_f}{P_0} = \frac{T_f}{T_0} \]
Given that \( P_f = 6P_0 \):
\[ 6 = \frac{T_f}{T_0} \]
This implies:
\[ T_f = 6T_0 \]

### Step 6: Solve for the Temperature Increase Factor
Now we have two expressions for \( T_f \):
1. \( T_f = 4.4 T_0 \)
2. \( T_f = 6 T_0 \)

Setting these equal gives:
\[ 4.4 T_0 = 6 T_0 \]
This indicates that the temperature has increased by a factor of:
\[ \frac{T_f}{T_0} = 6 \]

### Conclusion
The temperature of the gas increases by a factor of **6**.