91Ó°ÊÓ

An ideal-gas process is described by $\mathrm{ÒÏ}=c{V}^{1/2}$

Volume changes from$V_1$to$V_2$

(a) The work is equal to minus the area under the $p-V$diagram. As an integral, we can express it to be

$W=-{∫}_{V_1}^{V_2}pdV$

Knowing the formula giving the pressure as a function of the volume, we are left to only integrate:

$W=-{∫}_{V_1}^{V_2}c{V}^{1/2}dV=-c{\left[\frac{2}{3}{V}^{3/2}\right]}_{V_1}^{V_2}=\frac{2c}{3}\left(V_1^{3/2}-V_2^{3/2}\right)$

Therefore, the expression for the work done on the gas in this process as the volume changes is$\frac{2c}{3}\left(V_1^{3/2}-V_2^{3/2}\right).$

An ideal-gas process is described by $\mathrm{ÒÏ}=cV$

Amount of gas$=0.033mol$

Temperature $={150}^{∘}C$

Using this process from$300c{m}^3$to$200c{m}^3$

(b) In this case the only thing we need to find is the constant $c$and then substitute.

First, let's take the function providing the pressure as a function of the volume:

First, let's take the function providing the pressure as a function of the volume:

$p=c{V}^{1/2}⇒c=\frac{p}{V^{1/2}}$

Now we can express the initial pressure $p_1$in terms of the initial volume $V_1$and initial temperature $T_1$, from the ideal gas law:

$p_1{V}_1=nR{T}_1⇒p_1=\frac{nR{T}_1}{V_1}$

Substituting in the case of the initial parameters, we find the constant to be

$c=\frac{nR{T}_1}{V_1^{3/2}}$

Numerically, disregarding the units, this constant will be

localid="1648553527828" role="math" $c=\frac{(0.033mol·8.314Jâ‹\dots K^{-1}mo{l}^{-1}·423K)}{{\left(3·{10}^{-4}{m}^3\right)}^{3/2}}=\underset{¯}{2.23·{10}^7}$

Substituting in our expression found for the work, we have

$W=\frac{(2×2.23×{10}^7)}{3}\left({\left(3×{10}^{-4}{m}^3\right)}^{3/2}-{\left(2×{10}^{-4}{m}^3\right)}^{3/2}\right)=35\mathrm{J}$

An ideal-gas process is described by $\mathrm{ÒÏ}=cV$

(c) Knowing the relation between the pressure and volume, we can write:

$p=c{V}^{1/2}⇒\frac{p_1}{p_2}=\frac{V_1}{V_2}$

We can use this to substitute in the ideal gas law:

$\frac{p_1{V}_1}{T_1}=\frac{p_2{V}_2}{T_2}⇒T_2=\frac{p_2{V}_2}{p_1{V}_1}{T}_1$

Since we have the ratio of pressures given by the ratio of volumes, we can now express the unknown temperature by the ratio of volumes and the known initial one, finding:

$T_2={\left(\frac{V_2}{V_1}\right)}^{3/2}{T}_1$

In our numerical case, we have $T_1={150}^{∘}\mathrm{C}=423\mathrm{K}$which means that the final temperature will be,

$T_2={\left(\frac{200m^3}{300m^3}\right)}^{3/2}×423K=230\mathrm{K}=−{40}^{∘}\mathrm{C}$

Therefore, the final temperature of the gas is$-{40}^{∘}C.$