Question

in figure, an ideal gas is brought from an initial state to a final state by two processes. initial and final states are in the same temperature. the initial volume and pressure are vi = 0.55 m³ and pi = 115 kpa, and the final values for these variables are vf = 0.15 m³ and pf = 422 kpa. how much energy is transferred thermally to the gas during process a and process b? pv = nrt pivi = nrt ti = pivi / nr tf = pfvf / nr tf - ti = (pfvf - pivi) / nr a. ⇒ φtot = φ1 + φ2 φ1 = δeis + w1 = 0e w1 = ∫ pdv = 0 φ2 = δesf + w2 = δe + pf(vf - vi) w2 = ∫ pdv = pf(vf - vi) φtot = (δeis + δesf) + pf(vf - vi) φtot(a) = δesf + pf(vf - vi) φtot(a) = 1/2 nr(tf - ti) + pf(vf - vi) φtot(b) = δeif + pi(vf - vi) φtot(b) = 1/2 nr(tf - ti) + pi(vf - vi)

Explanation:

Step1: Recall the first - law of thermodynamics

$\Delta U = Q - W$, for an ideal gas with constant temperature $\Delta U=0$, so $Q = W$.

Step2: Calculate work for process A

For process A, the work done $W_A=\int PdV$. The process is a two - step process: first isobaric expansion and then isochoric process. The isochoric process does no work ($W_{isochoric}=0$). The isobaric work $W_{isobaric}=P_f(V_f - V_i)$. Given $P_i = 115\times10^{3}\ Pa$, $V_i = 0.55\ m^{3}$, $P_f = 422\times10^{3}\ Pa$, $V_f = 0.15\ m^{3}$. But we know for a constant - temperature process of an ideal gas, we can also use $W_A = nRT\ln\frac{V_f}{V_i}$. First, from $PV = nRT$, at constant $T$, $P_iV_i=P_fV_f$. The work done in process A: $W_A=P_iV_i\ln\frac{V_f}{V_i}=115\times10^{3}\times0.55\times\ln\frac{0.15}{0.55}$.
$W_A=115000\times0.55\times\ln(\frac{0.15}{0.55})\approx115000\times0.55\times(- 1.2528)\approx - 79997\ J$. The negative sign just indicates the direction of energy transfer. The magnitude of heat transfer $Q_A\approx80000\ J$.

Step3: Calculate work for process B

For process B, it is also a two - step process: first isochoric and then isobaric. The isochoric process does no work. The isobaric work $W_B = P_i(V_f - V_i)=115\times10^{3}\times(0.15 - 0.55)=115000\times(-0.4)= - 46000\ J$. The magnitude of heat transfer $Q_B = 46000\ J$.

Answer:

For process A, the heat transfer $Q_A\approx80000\ J$. For process B, the heat transfer $Q_B = 46000\ J$.