Question

calculate the change in internal energy if a system releases 25.5 j of heat to its surroundings while expanding a piston from 1.0 l to 1.8 l at a constant pressure of 1.05 atm.
fill in the blanks below based on this scenario.
the value for ( q ) is.
the value for ( delta v ) is.
determine values for ( q ) and ( p delta v ) for the reaction describe above.
( q=quad mathrm{j} )
( p delta v=quad mathrm{atm} mathrm{l} )
calculate ( delta e ) for this process. (note: ( 1 mathrm{~atm} mathrm{~l}=101.3 mathrm{~j} ))
( delta e=quad mathrm{j} )

Explanation:

Step1: Determine the value of \( q \)

The system releases heat, so \( q=- 25.5\ \text{J}\) (heat released is negative).

Step2: Calculate \(\Delta V\)

\(\Delta V=V_{final}-V_{initial}=1.8\ \text{L}-1.0\ \text{L} = 0.8\ \text{L}\)

Step3: Calculate \(P\Delta V\)

Given \(P = 1.05\ \text{atm}\) and \(\Delta V=0.8\ \text{L}\)
\(P\Delta V=1.05\ \text{atm}\times0.8\ \text{L}=0.84\ \text{atm L}\)

Step4: Convert \(P\Delta V\) to joules

Since \(1\ \text{atm L}=101.3\ \text{J}\), then \(P\Delta V = 0.84\times101.3\ \text{J}=85.092\ \text{J}\)

Step5: Calculate \(\Delta E\)

Using the first law of thermodynamics \(\Delta E=q - P\Delta V\)
Substitute \(q=-25.5\ \text{J}\) and \(P\Delta V = 85.092\ \text{J}\)
\(\Delta E=-25.5\ \text{J}-85.092\ \text{J}=-110.592\ \text{J}\approx - 110.6\ \text{J}\)

Answer:

\(q=-25.5\ \text{J}\)
\(P\Delta V = 0.84\ \text{atm L}\)
\(\Delta E=-110.6\ \text{J}\)