FIGURE P19.62 shows a thermodynamic process followed by $120\mathrm{mg}$of helium.

a. Determine the pressure (in $atm$), temperature (in ${}^{\circ }\mathrm{C}$), and volume in $\left(c{m}^3\right)$of the gas at points 1,2 , and 3 . Put your results in a table for easy reading.

b. How much work is done on the gas during each of the three segments?

c. How much heat energy is transferred to or from the gas during each of the three segments?

Amount of helium$=120mg$

Pressure$=?$

Temperature$=?$

Volume$=?$

(a) Let us first calculate $p_1$. From the Ideal Gas Law, and substituting $n=\frac{m}{M}$, we have

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

We can substitute numerically to find

$p_1=\frac{1.2·{10}^{-4}·8.314·(133+273)}{0.004·0.001}=\underset{¯}{1.01·{10}^5\mathrm{Pa}}.$

This is practically 1 atm.

Having determined $p_1$, we now know $p_2=5p_1=5$atm and $p_3=p_1=1\mathrm{atm}$.

Since the first process is an isochoric heating, the increase in temperature will be proportional to the increase in pressure . That is,

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

Numerically, we have

$T_2=5T_1=5·(133+273)=2030\mathrm{K}=\underset{¯}{1757°\mathrm{C}}$

The volume after the first process is still $\underset{¯}{V_2=V_1=1000{\mathrm{cm}}^3}$.

Since the second process is isothermic, the temperature after it will be

$\underset{¯}{T_3=T_2=1757°\mathrm{C}}$

We are now left with finding $V_3$. Since the second process is isothermic, we can say

$p_2{V}_2=p_3{V}_3\Rightarrow V_3=\frac{p_2}{p_3}{V}_2$

Knowing the results until now, we can find this volume to be

$\underset{¯}{V_3=5V_2=5V_1=5000{\mathrm{cm}}^3}$

Tabulated, our results are as follows:

Amount of helium

Amount of work is done on the gas during each of the three segments$=?$

(b) In the first process, no work is done on the gas.

The work on the isothermic process will be given by,

$W=-nRT\ln \left(\frac{V_3}{V_2}\right)$

In our case, we have

$W_2=-\frac{0.120}{4}·8.314·(1757+273)\ln \left(5\right)=-815\mathrm{J}$

In the isobaric process the work is given by

$W=p\Delta V$

Knowing the pressure and the two values of the volume, we have

$W_3=1·{10}^5·0.004=-400\mathrm{J}$

Amount of work is done on the gas during each of the three segments are,

$W_1=0,W_2=-815\mathrm{J},W_3=-400\mathrm{J}$

Amount of helium

Amount of heat energy is transferred to or from the gas during each of the three segments

$=?$

The heat taken in this isochoric process will be given by,

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

Knowing our values and that the number of moles can be given as $n=\frac{m}{M}$, we have

$Q_1=\frac{0.12}{4}·12.5·4·(1757-133)=2436\mathrm{J}$

The heat transferred in the second process will be, from the First Law, considering that there was no internal energy change, equal to the negative of the work. That is,

$Q_2=-W_2=815\mathrm{J}$

The heat exchanged in this isobaric compression will be given by

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

For our case, we will have

$Q_3=\frac{0.12}{4}·20.8·(1757-133)=1013\mathrm{J}$

Amount of heat energy is transferred to or from the gas during each of the three segments are,

$Q_1=2440\mathrm{J},Q_2=815\mathrm{J},Q_3=1013\mathrm{J}$