edited my answer, now the results are equal

1. If you re-calculated $$V_1$$ you'll find that it equals $$0.01503$$ not $$0.0158$$ $$m^3$$.

2. same issue with $$P_2$$. $$P_2 = \frac{138*10^{3}*0.01503^{1.4}}{(8.3529411*10^{-4})^{1.4}}= 7.889*10^6\ Pa$$

Substituting in isentropic work equation: $$W = \frac{P_2V_2 - P_1V_1}{k-1} = \frac{(7.867*10^6 * 8.3529411*10^{-4}) - (138*10^3*0.01503)}{1.4-1}=11.2 \ kJ$$

NOTE: I disagree with Andrea's answer, I'll explain later.

edited body
Algo
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1. If you re-calculated $$V_1$$ you'll find that it equals $$0.01503$$ not $$0.0158$$ $$m^3$$.

2. same issue with $$P_2$$. $$P_2 = \frac{138*10^{3}*0.01503^{1.4}}{(8.3529411*10^{-4})^{1.4}}= 7.889*10^6\ Pa$$

Substituting in isentropic work equation: $$W = \frac{P_2V_2 - P_1V_1}{k-1} = \frac{(7.867*10^6 * 8.3529411*10^{-4}) - (138*10^3*0.01503)}{1.4-1}=11.2 \ kJ$$

NOTE: I disagree with Andrea's answer, I'll explain later.

deleted 86 characters in body
Algo
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• 1
• 14
• 31

1. If you re-calculated $$V_1$$ you'll find that it equals $$0.01503$$ not $$0.0158$$ $$m^3$$.

2. same issue with $$P_2$$ is pressure after compression, logically it can't be lower than initial pressure, and that's number 2. $$P_2 = \frac{138*10^{3}*0.01503^{1.4}}{(8.3529411*10^{-4})^{1.4}}= 7.889*10^6\ Pa$$

Substituting in isentropic work equation: $$W = \frac{P_2V_2 - P_1V_1}{k-1} = \frac{(7.867*10^6 * 8.3529411*10^{-4}) - (138*10^3*0.01503)}{1.4-1}=11.2 \ kJ$$

NOTE: I disagree with Andrea's answer, I'll explain later.