Question

the air within a piston equipped with a cylinder absorbs 565 j of heat and expands from an initial volume of 0.10 l to a final volume of 0.85 l against an external pressure of 1.0 bar. what change in internal energy of the air within the piston?

Explanation:

Step1: Convert pressure unit

Since \(1\ bar = 10^{5}\ Pa\), so \(P = 1.0\ bar=1.0\times10^{5}\ Pa\)

Step2: Calculate volume change

\(\Delta V=V_{2}-V_{1}\), where \(V_{1} = 0.10\ L=0.10\times10^{- 3}\ m^{3}\), \(V_{2}=0.85\ L = 0.85\times10^{-3}\ m^{3}\)
\(\Delta V=(0.85 - 0.10)\times10^{-3}\ m^{3}=0.75\times10^{-3}\ m^{3}\)

Step3: Calculate work done

The formula for work done in expansion against constant - pressure is \(w=-P\Delta V\)
Substitute \(P = 1.0\times10^{5}\ Pa\) and \(\Delta V=0.75\times10^{-3}\ m^{3}\) into the formula
\(w=-1.0\times10^{5}\ Pa\times0.75\times10^{-3}\ m^{3}=- 75\ J\)

Step4: Calculate internal energy change

According to the first law of thermodynamics \(\Delta U=q + w\)
Given \(q = 565\ J\) and \(w=-75\ J\)
\(\Delta U=565\ J+( - 75\ J)=490\ J\)

Answer:

The work done \(w=-75\ J\) and the change in internal energy \(\Delta U = 490\ J\)