As a result of the isobaric change by 80 K, one mole of a certain ideal gas obtain an amount of heat equal to 1.60 kJ.
What is change in enthalpy of gas

Options

A. 0
B. 20 kJ
C. `-1.60` kJ
D. `+1.60` kJ

Answer

To solve the problem, we need to determine the change in enthalpy (ΔH) for one mole of an ideal gas undergoing an isobaric (constant pressure) process with a temperature change of 80 K and an absorbed heat of 1.60 kJ.

### Step-by-step Solution:

1. **Understand the Isobaric Process**:
In an isobaric process, the pressure remains constant. The heat absorbed by the system (q) is equal to the change in enthalpy (ΔH) when the process occurs at constant pressure.

2. **Identify Given Data**:
- Heat absorbed (q) = 1.60 kJ = 1.60 × 10³ J
- Temperature change (ΔT) = 80 K
- Number of moles (n) = 1 mole

3. **Apply the Relationship for Enthalpy Change**:
For an isobaric process, the change in enthalpy (ΔH) can be directly equated to the heat absorbed:
\[ \Delta H = q \]

4. **Substitute the Given Value**:
Since we know the heat absorbed (q) is 1.60 kJ, we can substitute this value into the equation:
\[ \Delta H = 1.60 \text{ kJ} \]

5. **Final Answer**:
Thus, the change in enthalpy (ΔH) for the gas is:
\[ \Delta H = 1.60 \text{ kJ} \]

### Conclusion:
The change in enthalpy of the gas is 1.60 kJ.

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