91Ó°ÊÓ

Diameter of the cylinder is $=10cm$

Pressure $=10atm$

Cylinder Initial length$=20cm$

Temperature$=2500J$

(a) Since the process is isothermic, the heat exchanged $Q$will be given as a function of the temperature change $∆T$by,

$Q=n{C}_p\mathrm{Δ}T$

Therefore, the temperature difference will be

$\mathrm{Δ}T=\frac{Q}{n{C}_p}$

Here we don't know the number $n$of moles, but again, we shouldn't calculate it numerically, but instead substitute its parametric expression.

From the ideal gas law, we can find $n$to be

$pV=nRT⇒n=\frac{pV}{RT}$

Surely enough, when substituting we have to substitute all $(p,V,T)$belonging to the same state.

Substituting, we can find the temperature difference to be

$\mathrm{Δ}T=\frac{Q}{C_p}\frac{RT}{pV}=\frac{QR{T}_1}{p_1{V}_1{C}_p}$

Therefore, the temperature $T_2$will be

$T_2=T_1+\mathrm{Δ}T=T_1\left(1+\frac{QR}{p_1{V}_1{C}_p}\right)$

If unconvinced, one can perform dimensional analysis to confirm our expression.

In our situation, we are given the diameter of the cylinder and the initial height, not the initial volume.

Therefore, we can write

$V_1=A{h}_1=\mathrm{Ï€}{R}^2{h}_1=\frac{\mathrm{Ï€}{D}^2}{4}{h}_1$

Substituting in the expression we found for the temperature, we have

$T_2=T_1\left(1+\frac{4QR}{\mathrm{Ï€}{p}_1{D}^2{h}_1{C}_p}\right)$

In our given numerical case, we will have

localid="1648560213496" $T_2=32K3\left(1+\frac{4×2500J×8.314Jâ‹\dots K^{-1}â‹\dots mo{l}^{-1}}{\mathrm{Ï€}×10atm×1.01·{10}^5·0.1^{2}cm·0.2m·20.8}\right)=526\mathrm{K}$

Therefore, the final temperature is$526k.$

Diameter of the cylinder is$=10\mathrm{cm}$

Pressure$=10\mathrm{atm}$

Cylinder Initial length$=20\mathrm{cm}$

Temperature$=2500\mathrm{J}$

Since the process is isobaric, we can write

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

The volume, however, as we mentioned, is $V=Ah$.

Substituting and then cancelling the cross-section areas we get

$\frac{A{h}_1}{T_1}=\frac{A{h}_2}{T_2}⇒h_2=\frac{T_2}{T_1}{h}_1$

In our numerical case, we will have

localid="1648567822414" $h_2=\left(\frac{526m^2}{423K}·20cm\right)=24.9\mathrm{cm}$

Therefore, final length of the cylinder is$24.9cm$