Your second attempt is correct, you just had two mistakes:

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

Which is very close to your problem set answer, I think it was a multiple choice question as he is asking about which answer nearly equals.

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

Your second attempt is correct, you just had two mistakes:

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

Which is very close to your problem set answer, I think it was a multiple choice question as he is asking about which answer nearly equals.

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

Your second attempt is correct, you just had two mistakes:

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

Which is very close to your problem set answer, I think it was a multiple choice question as he is asking about which answer nearly equals.

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Your second attempt is correct, you just had two mistakes:

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

Which is very close to your problem set answer, I think it was a multiple choosechoice question as he is asking about which answer nearly equals.

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

Your second attempt is correct, you just had two mistakes:

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

Which is very close to your problem set answer, I think it was a multiple choose question as he is asking about which answer nearly equals.

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

Your second attempt is correct, you just had two mistakes:

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

Which is very close to your problem set answer, I think it was a multiple choice question as he is asking about which answer nearly equals.

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

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Your second attempt is correct, you just had two mistakes:

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

Which is very close to your problem set answer, I think it was a multiple choose question as he is asking about which answer nearly equals.

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

Your second attempt is correct, you just had two mistakes:

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

  2. $P_2$ 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: $$W = \frac{P_2V_2 - P_1V_1}{k-1} = \frac{(7.867*10^6 * 8.3529411*10^{-4}) - (138*10^3*0.01503)}{1.4-1}=11.2 \ kJ$$

Which is very close to your problem set answer, I think it was a multiple choose question as he is asking about which answer nearly equals.

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

Your second attempt is correct, you just had two mistakes:

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

Which is very close to your problem set answer, I think it was a multiple choose question as he is asking about which answer nearly equals.

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

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