A cylinder-piston system contains an ideal gas at a pressure of [tex][tex]$1.5 \times 10^5 \, \text{Pa}$[/tex][/tex]. The piston is pushed out, allowing the gas to expand from an initial volume of [tex][tex]$0.0002 \, \text{m}^3$[/tex][/tex] to a final volume of [tex][tex]$0.0006 \, \text{m}^3$[/tex][/tex]. The system absorbs [tex][tex]$32 \, \text{J}$[/tex][/tex] of heat during this process. What is the change in internal energy? (Use [tex]\Delta U = Q - W = Q - P \Delta V[/tex])

A. [tex][tex]$-28 \, \text{J}$[/tex][/tex]
B. [tex][tex]$-92 \, \text{J}$[/tex][/tex]
C. [tex][tex]$92 \, \text{J}$[/tex][/tex]
D. [tex][tex]$28 \, \text{J}$[/tex][/tex]



Answer :

To solve this problem, we use the first law of thermodynamics, which states:

[tex]\[ \Delta U = Q - W \][/tex]

where [tex]\(\Delta U\)[/tex] is the change in internal energy, [tex]\(Q\)[/tex] is the heat absorbed by the system, and [tex]\(W\)[/tex] is the work done by the system.

Firstly, we need to calculate the change in volume ([tex]\(\Delta V\)[/tex]):

[tex]\[ \Delta V = V_{\text{final}} - V_{\text{initial}} \][/tex]
[tex]\[ \Delta V = 0.0006 \, m^3 - 0.0002 \, m^3 \][/tex]
[tex]\[ \Delta V = 0.0004 \, m^3 \][/tex]

Next, we calculate the work done by the gas during the expansion. For an ideal gas expanding at constant pressure, the work done ([tex]\(W\)[/tex]) is given by:

[tex]\[ W = P \Delta V \][/tex]

Substitute the given values for pressure ([tex]\(P = 1.5 \times 10^5 \, Pa\)[/tex]) and [tex]\(\Delta V\)[/tex]:

[tex]\[ W = 1.5 \times 10^5 \, Pa \times 0.0004 \, m^3 \][/tex]
[tex]\[ W = 60 \, J \][/tex]

Now, we apply the first law of thermodynamics. The system absorbs [tex]\(Q = 32 \, J\)[/tex] of heat. Therefore, the change in internal energy ([tex]\(\Delta U\)[/tex]) is:

[tex]\[ \Delta U = Q - W \][/tex]
[tex]\[ \Delta U = 32 \, J - 60 \, J \][/tex]
[tex]\[ \Delta U = -28 \, J \][/tex]

So, the change in internal energy is [tex]\(-28 \, J\)[/tex].

Thus, the correct answer is:
A. [tex]\( -28 \, J \)[/tex]

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