A Swiss watchmaker states in a flyer the mechanical power P_W dissipated by a specific clock (Fig. 2.1). The work provided to the clock is due to the temperature fluctuations ΔT around room temperature T that let the clock run during a time t. We consider that the atmospheric pressure p_{ext} is equal to the pressure of the gas p, i.e. p_{ ext} = p. The pressure p and the volume V of the gas are related by the ideal gas law p V = NR T where R is the ideal gas constant. Consider that the gas in the capsule is always at equilibrium with the air outside of the capsule (pressure and temperature are the same inside and outside). From the watchmaker’s information, estimate the volume V of the gas capsule used to run this clock.
Numerical Application:
P_W = 0.25 · 10^{−6} W, T = 25^◦C, ΔT = 1^◦C, p_{ ext} = 10^5 Pa and t = 1 day.
Question 2.6: A Swiss watchmaker states in a flyer the mechanical power PW...
According to relation (2.41), the infinitesimal work performed on the gas in the volume V inside the capsule, at constant atmospheric pressure p, is given by,
δW = P_W dt = −p (S, V) dV (reversible process) (2.41)
δ W = −p d V
The ideal gas law implies that,
d V =\frac{NR}{p} d T.
Thus, the infinitesimal work performed on the gas by the environment is given by,
δ W = −N R d T
The initial state i corresponds to the “hotter” equilibrium state just after a fluctuation and the final state f corresponds to the “colder” equilibrium state just before the next fluctuation. The work W_{if} performed on the gas between the initial state i, characterised by the thermodynamic quantities (V_i = V + ΔV, T_i = T + ΔT) and the final state ƒ, characterised by the thermodynamic quantities (V_ƒ = V, T_ƒ = T), is obtained when integrating the previous equation,
W_{if} = -\int_{T_i}^{T_f}{NRdT} = N R Δ T.
Moreover, the ideal gas law evaluated in the final state implies that,
W_{if} =\frac{p V}{T} ΔT .
Thus,
V = \frac{W_{if} T}{p ΔT} .
The work W_{if} must be equal to the product P_W t of the mechanical power P_W required to run the clock and the time t during which it runs, i.e. W_{if }= P_W t. Thus, the volume V of the capsule is given by,
V = \frac{P_W t T}{p ΔT} = 128 cm^3 .