Answer
a) $Q=−0.1 kJ$
b) $P_2=155.6\ kPa$
c) $\Delta L=7.1cm$
Work Step by Step
a) For an isothermal process, the energy balance becomes:
$Q−W_b=0$
Given $W_b=−0.1$: $Q=−0.1 kJ$
b) The initial volume is the volume of the 12cm diameter, 20cm long cylinder:
$V_1=\frac{\pi}{4}D^2L_1=0.002262 m³$
The expression of boundary work for isothermal processes is given by:
$W_b=P_1V_1ln(\frac{V_2}{V_1})$
Given $P_1=100\ kPa$, we can solve for the final volume: $V_2=0.001454 m³$
For ideal gases in isothermal processes:
$P_1V_1=P_2V_2\rightarrow P_2=155.6\ kPa$
c)The volume id given by:
$V_1=\frac{\pi}{4}D^2L_2$
$L_2=0.1285 m$
$ΔL=−0.07146m=-7.1cm$
