[Given: First law of thermodynamics $dU= dQ+dW$]
$dW= -pdV, dQ=0$
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.
The First Law of Thermodynamics states:
$dU = dQ + dW$
Where:
The problem provides two key conditions:
Substituting $dQ = 0$ into the First Law gives:
$dU = dW$
Substituting the expression for $dW$ gives:
$dU = -pdV$
The process is described as 'compression', meaning the volume of the air parcel decreases.
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.
The internal energy will increase.
If the work done on the system or by the system· is zero, which one of the following statements for a gas kept at a certain volume is correct?
A system that does NOT allow exchange of heat with its surrounding is called
A system that does NOT allow exchange of heat with its surrounding is called
For a certain reaction, ΔG θ = -45 kJ/mol and ΔH θ = -90 kJ/mol at 0 °C. What is the minimum temperature at which the reaction will become spontaneous, assuming that ΔH θ and ΔS θ are independent of temperature?