91Ó°ÊÓ

Each side of Cube $=20cm$

Amount of helium $=3.0g$

Temperature$={20}^{∘}C$

Heat Energy$=1000J$

Let us first make some notes, on what will be common in the solutions to both parts

(a) and (b):

Being given a mass $m$of material of molar mass $M$, the number $n$of moles will be

$n=\frac{m}{M}$

In both processes the temperature will change due to the heat given. Supposing molar specific heat $C_x,x∈\{p,V\}$, we have

$Q=n{C}_x\mathrm{Δ}T$

Therefore, the temperature raise will be

$\mathrm{Δ}T=\frac{Q}{n{C}_x}=\frac{QM}{m{C}_x}$

The temperature after the heat input therefore becomes

$T_2=T_1+\frac{QM}{m{C}_x}$

Finally, since we have a cube, the volume $V_1$will just be

$V_1=a_1^3,$

Where $a_1$is the side of the cube.

(a) Since the process is isochoric, we have

$\frac{p_1}{T_1}=\frac{p_2}{T_2}$

The pressure after the process finished can therefore be found as

$p_2=\frac{T_2}{T_1}{p}_1$

Knowing $T_2,$we have

$p_2=\frac{T_1+\frac{QM}{m{C}_v}}{T_1}{p}_1=\frac{\frac{T_{1}m{C}_v+QM}{m{C}_v}}{T_1}{p}_1=\frac{T_{1}m{C}_v+QM}{T_{1}m{C}_v}{p}_1$

Now we need to express the initial pressure in terms of known variables, which can be done through the Ideal Gas Law:

$p_1{V}_1=nR{T}_1=\frac{m}{M}R{T}_1⇒p_1=\frac{mR{T}_1}{M{V}_1}$

Substituting this, we have

$p_2=\frac{T_{1}m{C}_v+QM}{T_{1}m{C}_v}·\frac{mR{T}_1}{M{V}_1}$

Simplifying the mass, and temperature, we have

$p_2=\frac{\left(T_{1}m{C}_v+QM\right)R}{C_v{V}_{1}M}$

If one is not sure on the result, dimensional analysis proves this expression correct.

Numerically, for our case, we have

localid="1648615960119" role="math" $p_2=\left(\frac{({293}^{∘}C·0.003g·12.5m^3+1000J·0.004K/m)(8.314kgâ‹\dots m^2â‹\dots s^{-2}â‹\dots K^{-1})}{(12.5m^3×0.2m^3×0.004cm)}\right)$

$=3.12·{10}^6\mathrm{Pa}$

Hence, the final pressure if the process is at constant volume is$3.12·{10}^6\mathrm{Pa}$.

Each side of Cube$=20\mathrm{cm}$

Amount of helium$=3.0\mathrm{g}$

Temperature$=20°\mathrm{C}$

Heat Energy$=1000\mathrm{J}$

(b) Since the process is isobaric, we have

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

The final volume will therefore be

$V_2=\frac{T_2}{T_1}{V}_1=\frac{T_1+\frac{QM}{m{C}_p}}{T_1}{V}_1$

Finally,

$V_2=\frac{T_{1}m{C}_p+QM}{T_{1}m{C}_p}{V}_1$

For our numerical case, we have

localid="1648616102192" role="math" $V_2=\left(\frac{{293}^{∘}C·0.003g·20.8+1000J·0.004K/m}{{293}^{∘}C·0.2m^3·0.004cm}\right)$

$=7.3·{10}^{-3}{\mathrm{m}}^3$

Hence, the final volume if the process is at constant pressure is$7.3×{10}^{-3}{m}^3$.

Each side of Cube$=20\mathrm{cm}$

Amount of helium$=3.0\mathrm{g}$

Temperature$=20°\mathrm{C}$

Heat Energy$=1000\mathrm{J}$

The graph that we are looking for is the following:

Therefore, the diagram is as follows: