91Ó°ÊÓ

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

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

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,

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

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

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

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

$W=p\mathrm{Δ}V$

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,

$Q=n{C}_v\mathrm{Δ}T$

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

$Q=n{C}_p\mathrm{Δ}T$

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,

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