As a result of the isobaric heating by `DeltaT=72K`, one mole of a certain ideal gas obtain an amount of heat `Q=1.6kJ`. Find the work performed by the gas, the increment of its internal energy and `gamma`.

Explanation

To solve the problem step-by-step, we will follow the principles of thermodynamics, particularly focusing on the isobaric process for an ideal gas.

### Step 1: Calculate the Work Done by the Gas

In an isobaric process, the work done by the gas can be calculated using the formula:

\[ W = P \Delta V \]

However, we can also express this in terms of the change in temperature using the ideal gas law:

\[ W = nR \Delta T \]

Where:
- \( n \) = number of moles of the gas = 1 mole
- \( R \) = universal gas constant = 8.314 J/(mol·K)
- \( \Delta T \) = change in temperature = 72 K

Now substituting the values:

\[ W = 1 \times 8.314 \times 72 = 599.568 \text{ J} \approx 0.6 \text{ kJ} \]

### Step 2: Calculate the Change in Internal Energy

According to the first law of thermodynamics:

\[ \Delta U = Q - W \]

Where:
- \( Q \) = heat added to the system = 1.6 kJ
- \( W \) = work done by the gas = 0.6 kJ

Now substituting the values:

\[ \Delta U = 1.6 \text{ kJ} - 0.6 \text{ kJ} = 1.0 \text{ kJ} \]

### Step 3: Calculate the Ratio \( \gamma \)

The ratio \( \gamma \) (gamma) is defined as:

\[ \gamma = \frac{C_p}{C_v} \]

From the relationship between heat added and change in internal energy, we have:

\[ \frac{Q}{\Delta U} = \frac{nC_p \Delta T}{nC_v \Delta T} = \frac{C_p}{C_v} = \gamma \]

Substituting the values we have:

\[ \gamma = \frac{Q}{\Delta U} = \frac{1.6 \text{ kJ}}{1.0 \text{ kJ}} = 1.6 \]

### Final Results

1. Work done by the gas, \( W = 0.6 \text{ kJ} \)
2. Change in internal energy, \( \Delta U = 1.0 \text{ kJ} \)
3. Ratio \( \gamma = 1.6 \)