Question

$14g$ 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.

Short answer

a) The final temperature is $640K.$

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

c) The heat transfer to the gas is Zero.

d) The compression ratio is $8.498$

e) The diagrammatic representation of localid="1648101845261" $pV$diagram is

Step-by-step solution

Given Information (Part a)

Nitrogen gas at STP $=14g$

Pressure$=20atm$

The final temperature$=?$

Explanation (Part a)

(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

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

Final Answer (Part a)

Hence, the final temperature is$640K.$

Given Information (Part b) 

Nitrogen gas at STP

Pressure$=20atm$

The work done on the gas$=?$

Explanation (Part b) 

(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

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

Final Answer (Part b) 

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

Given Information (Part c) 

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

Pressure$=20\mathrm{atm}$

Heat transfer to the gas$=?$

Explanation (Part c) 

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

Final Answer (Part c) 

Hence, the heat transfer to the gas is Zero.

Given Information (Part d) 

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

Pressure $=20\mathrm{atm}$

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

Explanation (Part d)

(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$\frac{V_1}{V_2}=8.498$

Final Answer (Part d)

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

Given information (Part e)

Nitrogen gas at STP $=14g$

Pressure $=20atm$

Explanation (Part e)

Final Answer (Part e)  

A graphical representation of the process is given below.