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Question

An air parcel undergoes compression i.e its volume deceases, then its internal energy ____
[Given: First law of thermodynamics $dU= dQ+dW$]
$dW= -pdV, dQ=0$

The correct answer is
Will increase

Thermodynamics: Air Parcel Compression & Internal Energy

We need to determine the change in the internal energy of an air parcel undergoing compression, given the First Law of Thermodynamics and specific conditions.

Applying the First Law of Thermodynamics

The First Law of Thermodynamics states:

$dU = dQ + dW$

Where:

  • $dU$ is the change in internal energy.
  • $dQ$ is the heat added to the system.
  • $dW$ is the work done on the system.

Analyzing the Given Conditions

The problem provides two key conditions:

  1. Heat exchange is zero: $dQ = 0$.
  2. Work done during volume change: $dW = -pdV$.

Substituting $dQ = 0$ into the First Law gives:

$dU = dW$

Substituting the expression for $dW$ gives:

$dU = -pdV$

Evaluating the Change in Internal Energy

The process is described as 'compression', meaning the volume of the air parcel decreases.

  • This implies that the change in volume, $dV$, is negative ($dV < 0$).
  • Pressure ($p$) is always positive ($p > 0$).

Now, let's calculate the sign of $dU$:

$dU = -p \times dV$

$dU = -(positive) \times (negative)$

$dU = positive$

Since $dU$ is positive, the internal energy of the air parcel increases during compression when no heat is exchanged.

Conclusion

The internal energy will increase.

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