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Step 1: Recognize the Type of Process
The process is described as an isothermal expansion, meaning the temperature remains constant at 293 K.
Step 2: Apply the First Law of Thermodynamics
The first law states:
\Delta U = \Delta Q + W .
For an isothermal process involving an ideal gas, \Delta U = 0 , hence:
0 = \Delta Q + W ,
which gives
\Delta Q = -W .
Step 3: Express the Work Done
The expansion is carried out against a constant external pressure p_{\text{ext}} , so the work done by the system ( W ) is:
W = -p_{\text{ext}}\,(V_2 - V_1) ,
where V_1 and V_2 are initial and final volumes, respectively.
Step 4: Relate Volumes to Pressures for an Ideal Gas
For an ideal gas, pV = nRT . Thus:
V_1 = \frac{nRT}{p_1}
\quad \text{and} \quad
V_2 = \frac{nRT}{p_2}.
Step 5: Substitute into the Work Equation
Substituting the volumes in terms of p_1 and p_2 :
\[
W =
-\,p_{\text{ext}}
\biggl(
\frac{nRT}{p_2} - \frac{nRT}{p_1}
\biggr)
= -\,p_{\text{ext}}
\,nRT
\Bigl(
\frac{1}{p_2} - \frac{1}{p_1}
\Bigr).
\]
Step 6: Insert the Given Values and Calculate
Given:
p_{\text{ext}} = 4.3\,\text{MPa} ,
p_1 = 2.1\,\text{MPa} ,
p_2 = 1.3\,\text{MPa} ,
n = 5\,\text{mol} ,
T = 293\,\text{K} ,
and
R = 8.314\,\text{J mol}^{-1}\text{K}^{-1} .
First, express the work in Joules:
\[
W =
-\,4.3 \times 10^6\,\text{Pa}
\,\times 5\,\text{mol}
\,\times 8.314\,\text{J mol}^{-1}\text{K}^{-1}
\,\times 293\,\text{K}
\Bigl(
\frac{1}{1.3 \times 10^6\,\text{Pa}}
- \frac{1}{2.1 \times 10^6\,\text{Pa}}
\Bigr).
\]
Note that converting MPa to Pa involves multiplying by 10^6 .
Step 7: Interpret the Sign and Convert the Work to kJ mol−1
Because the process is expansion work against a larger external pressure, the work done by the system ( W ) is negative:
W \approx -15\,\text{kJ mol}^{-1} .
Therefore,
\Delta Q = -W \approx +15\,\text{kJ mol}^{-1}.
Step 8: State the Final Answer
The heat transferred in the process (per mole) is
+15\,\text{kJ mol}^{-1} (rounded to the nearest integer).
