A cube $20\mathrm{cm}$ on each side contains $3.0\mathrm{g}$ of helium at $20°\mathrm{C}$. $1000\mathrm{J}$ of heat energy are transferred to this gas. What are (a) the final pressure if the process is at constant volume and (b) the final volume if the process is at constant pressure? (c) Show and label both processes on a single $pV$ diagram.

Each side of Cube $=20cm$

Amount of helium $=3.0g$

Temperature$={20}^{\circ }C$

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 $C_x,x\in \{p,V\}$, we have

$Q=n{C}_x\Delta T$

Therefore, the temperature raise will be

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

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:

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

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" $p_2=\left(\frac{({293}^{\circ }C·0.003g·12.5m^3+1000J·0.004K/m)(8.314kg\cdot m^2\cdot s^{-2}\cdot K^{-1})}{(12.5m^3\times 0.2m^3\times 0.004cm)}\right)$

$=3.12·{10}^6\mathrm{Pa}$

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

$=7.3·{10}^{-3}{\mathrm{m}}^3$

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

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: