As a result of the isobaric heating by `DeltaT`=72 K one mole of a certain ideal gas obtains an amount of heat Q = 1.60 kJ. Find the work performed by the gas, the increment of its internal energy, and the value of `gamma=C_p//C_v`

Answer

To solve the problem step by step, we will use the principles of thermodynamics, particularly focusing on the isobaric process.

### Step 1: Understand the Given Data
- We have an isobaric process (constant pressure).
- The change in temperature, ΔT = 72 K.
- The amount of heat added, Q = 1.60 kJ = 1600 J.
- The number of moles, n = 1 mole.

### Step 2: Use the First Law of Thermodynamics
The first law of thermodynamics states:
\[ Q = \Delta U + W \]
Where:
- \( Q \) = heat added to the system
- \( \Delta U \) = change in internal energy
- \( W \) = work done by the gas

### Step 3: Calculate the Work Done by the Gas
For an isobaric process, the work done by the gas can be expressed as:
\[ W = P \Delta V \]
Using the ideal gas law, we can express \( W \) in terms of n, R, and ΔT:
\[ W = nR\Delta T \]
Substituting the values:
- \( n = 1 \) mole
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( \Delta T = 72 \, \text{K} \)

Calculating the work done:
\[ W = 1 \times 8.314 \times 72 \]
\[ W = 599.568 \, \text{J} \approx 600 \, \text{J} \]

### Step 4: Calculate the Change in Internal Energy
Now, we can rearrange the first law of thermodynamics to find \( \Delta U \):
\[ \Delta U = Q - W \]
Substituting the values:
\[ \Delta U = 1600 \, \text{J} - 600 \, \text{J} \]
\[ \Delta U = 1000 \, \text{J} \]

### Step 5: Calculate the Value of \( \gamma \)
We know the relation between specific heats:
\[ \gamma = \frac{C_p}{C_v} \]
For an isobaric process, we can express \( Q \) as:
\[ Q = nC_p\Delta T \]
Thus,
\[ C_p = \frac{Q}{n\Delta T} \]
Substituting the values:
\[ C_p = \frac{1600 \, \text{J}}{1 \times 72} \]
\[ C_p = 22.222 \, \text{J/(mol K)} \]

Using the relation \( C_p - C_v = R \), we can find \( C_v \):
\[ C_v = C_p - R \]
\[ C_v = 22.222 \, \text{J/(mol K)} - 8.314 \, \text{J/(mol K)} \]
\[ C_v = 13.908 \, \text{J/(mol K)} \]

Now, we can find \( \gamma \):
\[ \gamma = \frac{C_p}{C_v} = \frac{22.222}{13.908} \approx 1.598 \]

### Final Results
- Work done by the gas, \( W \approx 600 \, \text{J} \)
- Change in internal energy, \( \Delta U = 1000 \, \text{J} \)
- Value of \( \gamma \approx 1.6 \)