$14g$ of nitrogen gas at STP are adiabatically compressed to a pressure of $20\mathrm{atm}$. What are (a) the final temperature, (b) the work done on the gas, (c) the heat transfer to the gas, and (d) the compression ratio $V_{\max }/V_{\min }$? (e) Show the process on a $pV$ diagram, using proper scales on both axes.

Nitrogen gas at STP $=14g$

Pressure$=20atm$

The final temperature$=?$

(a) Let's take the Ideal Gas Law: applying it for the two states, before and after the expansion, we get

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

We know the ratio of the pressures, and will find the ratio of the volumes (for simplicity, we'll take $1\mathrm{atm}=1·{10}^5\mathrm{Pa}$, which is the STP pressure ; in fact, by definition, $1\mathrm{atm}=101300\mathrm{Pa}$.)

Since this is an adiabatic process, we have

$p{V}^{\gamma }=\text{const.}\Rightarrow p_1{V}_1^{\gamma }=p_2{V}_2^{\gamma }$

We can therefore say that the ratio of volumes will be

${\left(\frac{V_2}{V_1}\right)}^{\gamma }=\frac{p_1}{p_2}\Rightarrow \frac{V_2}{V_1}={\left(\frac{p_1}{p_2}\right)}^{1/\gamma }$

Substituting this in our previous result, we get

$T_2=\frac{p_2}{p_1}·{\left(\frac{p_1}{p_2}\right)}^{1/\gamma }{T}_1={\left(\frac{p_2}{p_1}\right)}^{\frac{\gamma -1}{\gamma }}{T}_1$

Finally, substituting the initial value of $T_1=273\mathrm{K}$and the other numerical values, we get

$T_2={20}^{\frac{1.4-1}{1.4}}·273=2.345·T_1=640\mathrm{K}$

Hence, the final temperature is$640K.$

Nitrogen gas at STP

Pressure$=20atm$

The work done on the gas$=?$

(b) In the adiabatic process, since there is no heat exchange, from the First Law, the work done will be equal to the change in internal energy .

That means that the work can be given as $W=n{C}_v\Delta T$

Since the change in internal energy is always has always the form as the work given above. Knowing that $T_2=2.345T_1$, we can easily find $\Delta T=1.345T_1=1.345·273=367\mathrm{K}$.

We also can find the number of moles as $n=\frac{m}{M}$.

Substituting, we have

$W=\frac{m}{M}{C}_v·\Delta T=\frac{14}{28}·20.8·367=3800\mathrm{J}$

The work done on the gas is$3800J.$

Nitrogen gas at STP $=14\mathrm{g}$

Pressure$=20\mathrm{atm}$

Heat transfer to the gas$=?$

By the definition of the adiabatic process, there is no heat exchange.

Hence, the heat transfer to the gas is Zero.

Nitrogen gas at STP$=14\mathrm{g}$

Pressure $=20\mathrm{atm}$

Compression ratiorole="math" localid="1648103265181" $V_{max/}{V}_{min}=?$

(d) The ratio of the volumes, relating to our simple derivations in part (a), can be found to be$\frac{V_2}{V_1}={\left(\frac{p_1}{p_2}\right)}^{1/V}$

Having already said that we'll take $p_1=p_a=1\mathrm{atm}$, we have $\frac{V_2}{V_1}={\left(\frac{1}{20}\right)}^{1/1.4}=\underset{¯}{1.4}\sqrt{0.05}=\underset{¯}{0.1177}$

In fact we're asked to calculate the opposite, the ratio between the highest and lowest volumes.

This will simply be the inverse, which is$\frac{V_1}{V_2}=8.498$

Therefore, the compression ratio$V_{max}/V_{min}=8.498$

Nitrogen gas at STP $=14g$

Pressure $=20atm$

A graphical representation of the process is given below.