One mole of an ideal gas is heated at constant pressure so that its temperature rises by `DeltaT`= 72 K. In The heat supplied is Q=1.6 kJ, find the change in its internal energy and the work done by the gas.

Answer

To solve the problem step by step, we will break it down into two parts: calculating the work done by the gas and then finding the change in internal energy.

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

1. **Identify the formula for work done at constant pressure (isobaric process)**:
\[ W = P \Delta V \]
However, we can use the ideal gas law to express this in terms of temperature change:
\[ W = nR \Delta T \]

2. **Substitute the known values**:
- Number of moles, \( n = 1 \) mole
- Ideal gas constant, \( R = 8.314 \, \text{J/(mol K)} \)
- Temperature change, \( \Delta T = 72 \, \text{K} \)

Now, substituting these values into the equation:
\[ W = 1 \times 8.314 \times 72 \]

3. **Calculate the work done**:
\[ W = 8.314 \times 72 = 598.608 \, \text{J} \]
Converting this to kilojoules:
\[ W = \frac{598.608}{1000} = 0.598608 \, \text{kJ} \approx 0.59 \, \text{kJ} \]

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

1. **Use the first law of thermodynamics**:
\[ Q = \Delta U + W \]
Rearranging gives:
\[ \Delta U = Q - W \]

2. **Substitute the known values**:
- Heat supplied, \( Q = 1.6 \, \text{kJ} \)
- Work done, \( W = 0.59 \, \text{kJ} \)

Now substituting these values:
\[ \Delta U = 1.6 - 0.59 \]

3. **Calculate the change in internal energy**:
\[ \Delta U = 1.6 - 0.59 = 1.01 \, \text{kJ} \]

### Final Results
- Work done by the gas, \( W \approx 0.59 \, \text{kJ} \)
- Change in internal energy, \( \Delta U \approx 1.01 \, \text{kJ} \)