One kg of air, initially at a temperature of 127°C, expands reversibly at a constant pressure until the volume is doubled. If the gas constant of air is 287 J/kg.K, the magnitude of work transfer is __________ kJ (round off to 2 decimal places).

The correct answer is 114.80.

Analysis of the Given Diagram:
The provided diagram illustrates a constant-pressure (isobaric) expansion process on a pressure-volume (P-V) plot. The path goes horizontally from state 1 to state 2, indicating that pressure remains constant:
$P=\text{constant}$
The image lists the following parameters for the system:
- Mass of air, $m=1\text{kg}$
- Initial temperature, $T_1=127°\text{C}=400\text{K}$
- Final volume is doubled, $V_2=2V_1$
- Gas constant of air, $R=287\text{J/(kg}·\text{K)}=0.287\text{kJ/(kg}·\text{K)}$

Step-by-Step Derivation and Calculation:

1. Convert the initial temperature from Celsius to Kelvin:
$$T_1=127+273=400\text{K}$$

2. For a reversible constant-pressure (isobaric) process, the work done ($W$) by the gas is given by:
$$W=P(V_2-V_1)$$

3. Since the volume doubles ($V_2=2V_1$), substitute $V_2$ into the work equation:
$$W=P(2V_1-V_1)=P{V}_1$$

4. Using the ideal gas equation of state at the initial state ($P{V}_1=mR{T}_1$), we can write the work done as:
$$W=mR{T}_1$$

5. Substitute the given values to compute the magnitude of the work transfer:
$$W=1\text{kg}\times 0.287\text{kJ/(kg}·\text{K)}\times 400\text{K}$$

$$W=114.80\text{kJ}$$