91Ó°ÊÓ

One cylinder's initial volume $=600c{m}^3$

Temperature$={30}^{∘}C$

Pressure$=1.0atm$

Heat Energy$=400J$

(a) Since the process involves an adiabatic change in energy, the work done on the gas will be equal to its internal energy. Given the internal energy change as an expression, we can write:

$n{C}_v\mathrm{Δ}T=\mathrm{Δ}E=W⇒n{C}_v\left(T_2-T_1\right)=W$

We can express the final temperature as,

$T_2=\frac{W}{n{C}_v}+T_1$

Clearly, we do not know how many moles there are.

Our understanding of this can be obtained from the ideal gas law applied in the initial instance, as follows,

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

Substituting, the final expression for the final temperature as

$T_2=\frac{WR{T}_1}{p_1{V}_1{C}_v}+T_1$

$T_2=T_1\left(\frac{WR}{p_1{V}_1{C}_v}+1\right)$

Substituting numerically, we have

$T_2=\left(\frac{400·8.314}{{10}^5·6·{10}^{-4}·20.8}+1\right)·303=1110\mathrm{K}$

Hence, the final temperature is$1110k.$

One cylinder's initial volume$=600{\mathrm{cm}}^3$

Temperature$=30°\mathrm{C}$

Pressure$=1.0\mathrm{atm}$

Heat Energy$=400\mathrm{J}$

(b) We know that these things hold true due to the adiabatic nature of the process:

$p_1{V}_1^{\mathrm{γ}}=p_2{V}_2^{\mathrm{γ}},$

which will allow us to express the needed volume as

localid="1648200519503" $V_2=V_1\mathrm{γ}\frac{p_1}{p_2}$

The ideal gas law can only predict that adiabatic processes take place in adiabatic processes

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

We can estimate the ratio of pressures using the following expression:

$\frac{p_1}{p_2}=\frac{V_2}{V_1}\frac{T_1}{T_2}$

Substituting, we have

$V_2=V_1\sqrt[2]{\frac{V_2}{V_1}\frac{T_1}{T_2}}$

Put $V_2$out of the root and $V_1$in, thus getting

$V_2=V_2^{\frac{1}{\mathrm{γ}}}\sqrt[\mathrm{γ}]{V_1^{\mathrm{γ}}\frac{1}{V_1}\frac{T_1}{T_2}}$

Combine $V_2$to one side and simplify $V_1$under the root,

$V_2^{\frac{\mathrm{γ}-1}{\mathrm{γ}}}={\left(V_1^{\mathrm{γ}-1}\frac{T_1}{T_2}\right)}^{\frac{1}{\mathrm{γ}}}$

Simplify the roots,

$V_2^{\mathrm{γ}-1}=V_1^{\mathrm{γ}-1}\frac{T_1}{T_2}$

Taking the $\mathrm{γ}-1th$root,

$V_2=V_1\sqrt[\mathrm{γ}-1]{\frac{T_1}{T_2}}$

From the result of part (a), the temperature ratio as

$T_2=T_1\left(\frac{WR}{p_1{V}_1{C}_v}+1\right)=T_1\frac{WR+p_1{V}_1{C}_v}{p_1{V}_1{C}_v}$

$\frac{T_1}{T_2}=\frac{p_1{V}_1{C}_v}{WR+p_1{V}_1{C}_v}$

The parametric solution is,

$V_2=V_1\sqrt[\mathrm{γ}-1]{\frac{p_1{V}_1{C}_v}{WR+p_1{V}_1{C}_v}}$

Since the expression contains no unknown variables, it is now straightforward to substitute and find,

$V_2=6·{10}^{-4}\sqrt[1.40-1]{\frac{{10}^5·6·{10}^{-4}·20.8}{400·8.314+{10}^5·6·{10}^{-4}·20.8}}=2.33×{10}^{-5}{m}^3$

In cubic centimeters,

$V_2=23{\mathrm{cm}}^3$

Hence, the volume is$23c{m}^3.$