Answer to Question #237768 in Molecular Physics | Thermodynamics for NANA

Question

Below is a 𝑇𝑉 diagram of a reversible process of a monatomic ideal gas with 𝑛 moles such that 𝑛𝑅 = 2.00 J K −1 . The process traces the curve 𝑇 = (400 K m−3 ) 𝑉 − (100 K m−6 ) 𝑉 2 starting at state 𝑎 with 𝑉𝑎 = 3.00 m3 and ending at state 𝑏 with with 𝑉𝑏 = 1.00 m3.Calculate, for this process (ensure that signed values have the correct sign): (a) Pressures 𝑃𝑎 and 𝑃𝑏 at the endpoints, in Pa (b) Change in internal energy, ∆𝑈, in J (c) Work 𝑊 done by the gas, in J (d) Heat 𝑄 added to the gas, in J (e) Entropy change Δ𝑆 of the gas, in J K.

Answer 1

(a) First, we use the ideal gas law and the equation that relates temperature and volume to find the pressure of the system:

$PV=nRT \implies P=\cfrac{nRT}{V} \\P=(nR)\cfrac{T}{V}=(nR)\cfrac{(AV+BV^2)}{V} \\P=(2 \frac{J}{K})(400 \frac{K}{m^3}-100\frac{K}{m^6}V)$

Then we substitute to find P_a and P_b:

$\\P_a=(2 \frac{J}{K})(400 \frac{K}{m^3}-100\frac{K}{m^6} (3\,m^3) ) \\ P_a=(2)(400-300)\,Pa=200\,Pa \\P_b=(2 \frac{J}{K})(400 \frac{K}{m^3}-100\frac{K}{m^6}(1\,m^3)) \\ P_b=(2)(400-100)\,Pa=600\,Pa$

We also know from the equation that relates volume and temperature that:

$T=f(V)=(400 \frac{K}{m^3})V-(100\frac{K}{m^6})V^2 \\ \implies dT= \Big[ (400 \frac{K}{m^3})-(100\frac{K}{m^6})V \Big]dV$

With that information we proceed to calculate the work done on this process and the change on the internal energy:

$W=\int pdV= \int nRdT \\ W=nR \int_{V_a}^{V_b} \Big[ (400 \frac{K}{m^3})- (100\frac{K}{m^6}) V \Big]dV \\ W=(2 \frac{J}{K}) \Big[ (400 \frac{K}{m^3}) (V_b-V_a) - (50\frac{K}{m^6}) (V^2_b-V^2_a) \Big] \\ W=\Big[ (800 \frac{J}{m^3}) (1-3)m^3 - (100\frac{J}{m^6}) ((1)^2-(3)^2)m^6\Big] \\ \implies W= [(800)(-2)-(100)(-8)]\,J=[-1600+800] \,J=-800\,J$

Now, to determine the change in internal energy we use $\Delta U=nC_vdT$, knowing that for a monoatomic gas C_v=3/2R:

$\Delta U=\int nC_vdT=\int \frac{3}{2}nRdT \\ \Delta U= \frac{3}{2}nR \int_{V_a}^{V_b} \Big[ (400 \frac{K}{m^3})-(100\frac{K}{m^6})V \Big]dV \\ \Delta U= \frac{3}{2}(2 \frac{J}{K}) \Big[ (400 \frac{K}{m^3}) (V_b-V_a) - (50\frac{K}{m^6}) (V^2_b-V^2_a) \Big] \\ \Delta U= (1200 \frac{J}{m^3}) (1-3)m^3 - (75\frac{J}{m^6}) ((1)^2-(3)^2)m^6 \\ \implies \Delta U=[(1200)(-2)-(75)(-8)]\,J=[-2400+600]\,J=-1800\,J$

With that information we use the definition for the internal energy change to find the heat change:

$\text{ } \Delta U=Q-W \implies Q=\Delta U + W \\Q= (-1800-800)\,J=-2600\,J$

Then, since the temperature for the initial and final states is the same (because T_a = T_b = 300 K) we proceed to calculate $\Delta S$ :

$\Delta S=\cfrac{Q}{T} =\cfrac{-2600\,J}{300\,K}=-8.67\cfrac{J}{K}$

In conclusion:

(a) Pressures 𝑃𝑎 and 𝑃𝑏 at the endpoints, in Pa: P_a = 200 Pa and P_b = 600 Pa.(b) Change in internal energy, ∆𝑈, in J: ∆𝑈= - 1800 J(c) Work 𝑊 done by the gas, in J: 𝑊 = - 800 J(d) Heat 𝑄 added to the gas, in J: 𝑄 = - 2600 J(e) Entropy change Δ𝑆 of the gas, in J/K: Δ𝑆 = -8.67 J/K

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.