1. The N moles of gas contained in the ith cylinder satisfy the ideal gas equation of state (5.47), i.e.
pV = NR T
p_i V_i = NR T.
The norm F_i exerted on the lever by the N moles of gas contained in the ith cylinder is,
F_i=\left\|F_i\right\| =p_iA=\frac{NR TA}{V_i}.
2. The mechanical condition implied by the equilibrium is that the net torque M^{ext}_{i} with respect to the fulcrum of the lever must vanish. When the lever is horizontal, this condition implies,
\sum\limits_{i=1}^{3}{M^{ext}_i} =0 . ⇒ (2F_1 + F_2 − 3F_3) d = 0.
Dividing this condition by Ad, we obtain a relation between the pressures in each cylinder, i.e.
2p_1 + p_2 − 3p_3 = 0.
3. The infinitesimal variations dV_i of the volume of gas in the ith cylinder is given by,
dV_i = Adh_i .
where dh_i is the height variation and A is the cross-section of the piston. The data on the figure imply that,
dh_1 = 2dh_2 and dh_3 = −3dh_2 .
which in turn implies that after multiplication by the cross-section A,
dV_1 = 2dV_2 and dV_3 = −3dV_2 .
Therefore, the total volume of gas V is constant, i.e.
dV_1 + dV_2 + dV_3 = 0 and thus V = V_1 + V_2 + V_3 = const.
4. Taking into account the extensive character of the internal energy, during an infinitesimal isothermal process (i.e. dT = 0), the internal energy variation dU_i of the gas contained in the ith cylinder vanishes. This implies that the infinitesimal internal energy variation (5.62) of the system vanishes, i.e.
U = cNRT
dU_i = cNRdT = 0 and thusdU=\sum\limits_{i=1}^{3}{dU_i} =0.
5. Taking into account the Gibbs relation (4.1) and the extensive characters of entropy and volume during an infinitesimal isothermal process associated with the motion of the lever, the infinitesimal entropy variation dS of the system is given by,
dU=TdS-pdV+\sum\limits_{A=1}^{r}{\mu _AdN_A}.
dS=\sum\limits_{i=1}^{3}{dS_i}=\sum\limits_{i=1}^{3}{\frac{p_idV_i}{T} }=NR\sum\limits_{i=1}^{3}{\frac{dV_i}{V_i} }=NR\biggl(\frac{dV_1}{V_1} +\frac{dV_2}{V_2}+\frac{dV_3}{V_3}\biggr).
Thus, we find that the energy variation of the system during an infinitesimal change in position of the lever vanishes, i.e.
dS=NR dV_2 \biggl(\frac{2}{V_1} +\frac{1}{V_2}+\frac{3}{V_3}\biggr)=\frac{dV_2}{T}\biggl(2p_1+p_2+3p_3\biggr)=0.
