IPhO 2022 · Theory problem 2

James Webb Space Telescope

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This is a question on the physics of the James Webb Space Telescope. Light from a star strikes the primary mirror, with an area of \(A_{\mathrm{mirror}}=25\,\mathrm{m^2}\), and reflects off of a secondary mirror. The focal length of the system is \(f=130\,\mathrm m\). The light is focused into the ISIM (Integrated Science Instrument Module), which contains the CCD (charged-coupled device) cameras.

Figure 1

Image Credit: NASA

Part A. Imaging a Star (1.8 points)

The nearest Red Giant is 89 light-years distant, has a temperature of \(T_{\mathrm{star}}=3600\,\mathrm K\), and a diameter of \(d_o=1.7\times10^{11}\,\mathrm m\).

A.1 — 0.4pt

Calculate the diameter of a focused image of the star on the CCD camera imaging surface.

A.2 — 0.4pt

Estimate the diameter of a diffraction central maximum on the CCD camera imaging surface. Assume a wavelength of \(\lambda=800\,\mathrm{nm}\), which is the strongest intensity wavelength from the red giant star.

A.3 — 1.0pt

If the CCD is not cooled and can lose heat only by radiating from the top of the imaging surface, what would be the equilibrium temperature of the CCD at the location of the image of the red giant star? Assume the CCD surface is a blackbody. Provide a formula and a numerical estimate.

Part B. Counting Photons (1.8 points)

The absorption of a photon by the CCD camera leads to the emission of an electron within the apparatus. This occurs only if the photon has sufficient energy to excite an electron across an energy gap \(\Delta E_g\). Assume that every photon with sufficient energy succeeds. There is also leakage of electrons across the gap caused by the temperature of the CCD camera; this is the dark current \(i_d\) and is measured in the number of electrons per second. It is a function of temperature according to

\[ i_d=i_0e^{-|\Delta E_g|/(6k_BT)}. \tag{1} \]

where \(i_0\) is a constant.

Figure 2

The graph shows how dark current varies with temperature. The units for dark current, \(\mathrm{e^-}/\mathrm s\) should be thought of as counting a number of electrons per second.

B.1 — 0.4pt

From the dark current graph, provide an order of magnitude estimate for the temperature of a distant source of thermal photons that would just be capable of exciting an electron on the pixel.

The electrons are collected in a capacitor, and after an exposure time \(\tau\), the electrons are counted. There are three main sources of uncertainty in the process: a fixed uncertainty in the counting process called read out noise; a Poisson distribution error associated with the dark current, and a Poisson distribution error associated with the detected incoming photons. Poisson distribution errors are equal to the square root of the number of counts associated with a process. The measured photon count is equal to the number of electrons in the capacitor, minus the number of electrons associated with the dark current.

B.2 — 0.4pt

Write an expression for the total count uncertainty \(\sigma_t\), if there is a readout noise \(\sigma_r\), a dark current \(i_d\), an incoming photon rate \(p\), and an exposure time \(\tau\).

For remaining questions in this part assume the exposure time is \(\tau=10^4\,\mathrm s\) and the read out noise is a fixed \(\sigma_r=14\).

B.3 — 0.5pt

Assume an operating temperature of \(T_p=7.5\,\mathrm K\). Calculate the minimum photon rate \(p\) so that the photon count is ten times the count uncertainty.

B.4 — 0.5pt

Assuming all photons are just capable of exciting an electron across the band gap, what is the intensity of the source of photons found in B.3 on the primary mirror? Express your answer in \(\mathrm{W/m^2}\)

Part C. Passive Cooling (4.4 points)

An infrared CCD camera must be kept at a low temperature. The first tool is a shield to protect from the sun's radiation.

The sun-shield consists of five separated reflective layers in thin sheets (black); radiant energy (gray) from the sun is incident on the first sheet on left, and some energy escapes between every pair of sheets.

Figure 3

Schematic of energy flow: the vertical lines (black) are the sheets, the flow of energy (gray) is from the left to the right, however, between sheets, some energy flows up and out into space.

Figure 4

On the left is a simple model of two adjacent sheets 1 and 2 separated by a distance \(h\). The sheets are not connected, and the perimeter is open to space. Assume the sheets are parallel. Thermal radiation can be exchanged between the sheets, and thermal radiation can escape through the perimeter gap. On the right, the perimeter gap has been shaded to help visualize.

Assume the following simplifications:

  • Sheets are square, each with area \(A_{\mathrm{sheet}}=200\,\mathrm{m^2}\).
  • Sheets are parallel and separated by \(h=25\,\mathrm{cm}\) along the perimeter.
  • Sheets have constant emissivity \(\epsilon\ll1\). Assume that all reflections off of sheet surfaces are diffuse.
  • Sheets are thin with temperature on the front and back surfaces equal and uniform.
  • The fraction of radiant flux emitted by a sheet that is absorbed by the adjacent sheet is \(\alpha\leq1\). This means that if sheet 1 in the figure above emits an amount of heat \(Q_1\) toward sheet 2 then sheet 2 will absorb an amount \(\alpha Q_1\) from sheet 1.
  • The amount of radiant flux ejected out of the perimeter gap between two sheets is approximated as \(\beta Q_{12}\) where \(\alpha Q_{12}\) is the net flux between the two sheets. The fraction \(\beta<1\). This is equivalent to saying that the heat loss to space between two sheets is proportional to the net heat exchange between the sheets. This is a rough approximation for this problem.
  • Background temperature of space is negligible.

C.1 — 2.4pt

Derive expressions for the equilibrium temperatures of the first sheet and fifth sheet in terms of the incident solar radiation intensity \(I_0\), the constants \(\alpha\) and \(\beta\), and any necessary physical constants. To simplify your expression, you may define additional constants in terms of \(\alpha\) and \(\beta\), etc.

C.2 — 1.6pt

Derive numerical estimates for \(\alpha\) and \(\beta\) from the information about the sheet geometry assuming an emissivity \(\epsilon=0.05\). You are encouraged to consider the rectangular box model of the sheets above, where the perimeter area effectively acts as a perfect absorber of radiant energy.

C.3 — 0.4pt

Numerically determine the temperatures of sheet 1 and sheet 5. The solar intensity is \(I_0=1360\,\mathrm{W/m^2}\).

Part D. Cryo-cooler (4 points)

The last stage of the cooling system directly cools the CCD camera. A closed cycle refrigeration system has a supply pipe line feeding helium gas at constant pressure \(P_1\) moving through a sponge like porous plug into a pipe with constant pressure \(P_2\). The pipe carries the gas to cool the CCD. The helium gas then passes through a pump before returning to the supply line.

Figure 5

Helium gas supplied on the left at well defined pressure \(P_1\) and temperature \(T_1\) is forced through the plug to well defined pressure \(P_2\) and temperature \(T_2\), where it is carried away on the right.

As the gas moves through the porous plug, viscous friction with the narrow walls of the channels in the sponge becomes an important effect; however, no heat is transferred to or from the gas during the process. The bulk speed of the gas in region 2 is only marginally greater than the bulk speed in region 1.

Helium is not an ideal gas, but does remain in a gaseous state throughout this process.

D.1 — 1.0pt

Consider a mole of gas that passes from left to right through the plug. Complete the table in your answer sheet by writing '>' or '<' to identify the quantity that must be greater, '=' to identify quantities that must be equal, or '?' if it is not possible to know which is greater or equal without more information.

Figure 6

D.2 — 0.6pt

Identify a conserved quantity constructed from \(U\) (internal energy), \(P\) (pressure), and \(V\) (volume) as a mole of gas moves through the plug; show work on how you derived this conserved quantity.

Your answer sheets have graphs of internal energy per mass against volume per mass for helium with isotherms and lines of constant entropy.

D.3 — 1.4pt

Assuming that \(V_2=0.100\,\mathrm{m^3/kg}\) and \(T_2=7.5\,\mathrm K\), use the graph to find a numerical value for the conserved quantity that you found in Part D.2. Show the construction on the graph!

Figure 7

D.4 — 0.8pt

Find the maximum possible temperature for \(T_1\). Show the construction on the graph!

Figure 8

D.5 — 0.2pt

Assuming your value for the maximum \(T_1\) found in D.4, find a numerical value for \(P_1\).

Physical Constants

Below is the list of physical constants you might use in your solution in addition to the quantities given in the problem text.

Physical constant Notation Numerical value
Speed of light in vacuum \(c\) \(299\,792\,458\,\mathrm{m/s}\)
Vacuum permeability (magnetic constant) \(\mu_0\) \(4\pi\cdot10^{-7}\,\mathrm{N/A^2};\quad[1\,\mathrm{N/A^2}=1\,\mathrm{V\,s/(A\,m)}]\)
Vacuum permittivity (electric constant) \(\varepsilon_0\) \(8.8541878128(13)\cdot10^{-12}\,\mathrm{F/m};\quad[1\,\mathrm{F/m}=1\,\mathrm{A\,s/(V\,m)}]\)
Elementary charge \(e\) \(1.602176634\cdot10^{-19}\,\mathrm C;\quad[1\,\mathrm C=1\,\mathrm{A\cdot s}]\)
Electron rest mass \(m_e\) \(9.1093837015(28)\cdot10^{-31}\,\mathrm{kg}\)
Proton rest mass \(m_p\) \(1.67262192369(51)\cdot10^{-27}\,\mathrm{kg}\)
Neutron rest mass \(m_n\) \(1.67492749804(95)\cdot10^{-27}\,\mathrm{kg}\)
Atomic mass unit \(m_{\mathrm{amu}}\) \(1.66053906660(50)\cdot10^{-27}\,\mathrm{kg}\)
Gravitational constant \(G\) \(6.67430(15)\cdot10^{-11}\,\mathrm{m^3/(kg\cdot s^2)}\)
Avogadro constant \(N_A\) \(6.02214076\cdot10^{23}\,\mathrm{mol^{-1}}\)
Molar gas constant \(R\) \(8.31446261815324\ldots\,\mathrm{J/(K\cdot mol)}\)
Boltzmann constant \(k_B\) \(1.380649\cdot10^{-23}\,\mathrm{J/K}\)
Stefan–Boltzmann constant \(\sigma\) \(5.670374419\ldots\cdot10^{-8}\,\mathrm{W/(m^2\cdot K^4)}\)
Planck constant \(h\) \(6.62607015\cdot10^{-34}\,\mathrm{J\cdot s}\)
Reduced Planck constant \(\hbar=h/(2\pi)\) \(1.0545718176\ldots\cdot10^{-34}\,\mathrm{J\cdot s}\)

Official English text of the problem.

Answers

  • A.1: \(d_{\mathrm{image}}\simeq2.6\times10^{-5}\,\mathrm m=26\,\mu\mathrm m\).
  • A.2: \(d_{\mathrm{diff}}\simeq4.5\times10^{-5}\,\mathrm m=45\,\mu\mathrm m\), using an equivalent circular aperture.
  • A.3: With distance \(L\) and a uniformly heated diffraction spot, \[ T_{\mathrm{image}}\simeq T_{\mathrm{star}} \left(\frac{A_{\mathrm{mirror}}d_o^2} {\pi L^2d_{\mathrm{diff}}^2}\right)^{1/4} \simeq4.1\times10^2\,\mathrm K. \]
  • B.1: The characteristic-energy estimate is \(T_{\mathrm{source}}\sim10^3\,\mathrm K\), since \(|\Delta E_g|/k_B\simeq1.4\times10^3\,\mathrm K\). A wavelength-peak criterion instead gives approximately \(2.8\times10^2\,\mathrm K\).
  • B.2: \(\displaystyle \sigma_t=\sqrt{\sigma_r^2+i_d\tau+p\tau}\).
  • B.3: \(\displaystyle p_{\min}\simeq0.22\,\mathrm{s^{-1}}\).
  • B.4: \(\displaystyle I\simeq1.7\times10^{-22}\,\mathrm{W/m^2}\).
  • C.1: Define \[ r=1+\frac{\beta}{\alpha}, \qquad K=1+\frac{1+r+r^2+r^3}{\alpha}. \] Then \[ T_1=\left[\frac{I_0}{\sigma}\frac{K}{K+r^4}\right]^{1/4}, \qquad T_5=\left[\frac{I_0}{\sigma(K+r^4)}\right]^{1/4}. \]
  • C.2: \(\alpha\simeq0.293,\quad\beta\simeq0.414\).
  • C.3: \(T_1\simeq360\,\mathrm K,\quad T_5\simeq120\,\mathrm K\).
  • D.1: In table order: \(?,\,?,\,<,\,>,\,<\).
  • D.2: \(\displaystyle H=U+PV\), neglecting the small change in bulk kinetic energy.
  • D.3: \(\displaystyle H/m\simeq4.0\times10^4\,\mathrm{J/kg}\), equivalent to approximately \(160\,\mathrm{J/mol}\) for helium.
  • D.4: \(\displaystyle T_{1,\max}\simeq11\,\mathrm K\).
  • D.5: \(\displaystyle P_1\simeq2\times10^6\,\mathrm{Pa}\).

Solution

A.1

Let the star’s distance be \(L\). Using \(1\) light-year \(\simeq9.46\times10^{15}\,\mathrm m\),

\[ L\simeq89(9.46\times10^{15}) =8.42\times10^{17}\,\mathrm m. \]

Since \(L\gg f\), the image lies essentially in the focal plane. Its angular diameter is \(d_o/L\), so

\[ d_{\mathrm{image}}=f\frac{d_o}{L} =\frac{130(1.7\times10^{11})}{8.42\times10^{17}} =2.62\times10^{-5}\,\mathrm m. \]

Thus

\[ \boxed{d_{\mathrm{image}}\simeq26\,\mu\mathrm m.} \]

A.2

Approximate the primary mirror by a circular aperture of the same area:

\[ D=2\sqrt{\frac{A_{\mathrm{mirror}}}{\pi}} =5.64\,\mathrm m. \]

For a circular aperture, the first diffraction minimum has angular radius \(1.22\lambda/D\). Consequently, the central maximum has diameter

\[ d_{\mathrm{diff}}=2.44\frac{\lambda f}{D}. \]

Numerically,

\[ d_{\mathrm{diff}} =2.44\frac{(800\times10^{-9})(130)}{5.64} =4.50\times10^{-5}\,\mathrm m. \]

Therefore,

\[ \boxed{d_{\mathrm{diff}}\simeq45\,\mu\mathrm m.} \]

The diffraction diameter exceeds the geometrical image diameter.

A.3

Treat the star as a blackbody and neglect optical losses. Its luminosity is

\[ P_{\mathrm{star}}=4\pi\left(\frac{d_o}{2}\right)^2 \sigma T_{\mathrm{star}}^4. \]

The power collected by the mirror is therefore

\[ P_{\mathrm{collected}} =A_{\mathrm{mirror}}\frac{P_{\mathrm{star}}}{4\pi L^2} =A_{\mathrm{mirror}}\sigma T_{\mathrm{star}}^4 \frac{d_o^2}{4L^2} \simeq2.43\times10^{-6}\,\mathrm W. \]

For the requested estimate, take the illuminated spot diameter to be the larger of the geometrical and diffraction diameters:

\[ d_{\mathrm{spot}}\simeq d_{\mathrm{diff}}, \qquad A_{\mathrm{spot}}\simeq\frac{\pi d_{\mathrm{diff}}^2}{4}. \]

Approximate the heating as uniform over this spot. Since only the top surface radiates, equilibrium requires

\[ P_{\mathrm{collected}} =\sigma A_{\mathrm{spot}}T_{\mathrm{image}}^4. \]

Thus

\[ \boxed{ T_{\mathrm{image}}\simeq T_{\mathrm{star}} \left( \frac{A_{\mathrm{mirror}}d_o^2} {\pi L^2d_{\mathrm{diff}}^2} \right)^{1/4} } \]

and

\[ \boxed{T_{\mathrm{image}}\simeq405\,\mathrm K \approx4.1\times10^2\,\mathrm K.} \]

This estimate neglects the detailed diffraction profile and the energy in its outer rings. In the geometrical-optics limit, substituting \(d_{\mathrm{spot}}=fd_o/L\) would instead give

\[ T_{\mathrm{image,geom}} =T_{\mathrm{star}} \left(\frac{A_{\mathrm{mirror}}}{\pi f^2}\right)^{1/4} \simeq530\,\mathrm K. \]

B.1

Taking logarithms of the dark-current law gives

\[ \ln i_d=\ln i_0-\frac{|\Delta E_g|}{6k_B}\frac1T. \]

From the graph, increasing \(1/T\) from \(0.12\,\mathrm{K^{-1}}\) to \(0.16\,\mathrm{K^{-1}}\) reduces the current from \(10^2\) to \(10^{-2}\,\mathrm{s^{-1}}\). Hence

\[ \frac{\Delta\ln i_d}{\Delta(1/T)} =\frac{\ln(10^{-4})}{0.04\,\mathrm{K^{-1}}} =-100\ln10\,\mathrm K. \]

It follows that

\[ \frac{|\Delta E_g|}{k_B} =600\ln10\,\mathrm K \simeq1.38\times10^3\,\mathrm K, \]

or

\[ \boxed{|\Delta E_g|\simeq1.91\times10^{-20}\,\mathrm J \simeq0.119\,\mathrm{eV}.} \]

Using the characteristic thermal-energy estimate \(E_\gamma\sim k_BT\),

\[ \boxed{T_{\mathrm{source}}\sim10^3\,\mathrm K.} \]

There is a convention-dependent numerical factor here. The threshold wavelength is

\[ \lambda_g=\frac{hc}{|\Delta E_g|} \simeq10.4\,\mu\mathrm m. \]

If “just capable” means that the wavelength peak of the thermal spectrum reaches this threshold, Wien scaling and the star’s given peak yield

\[ T_{\mathrm{source}} \simeq3600\,\mathrm K\, \frac{0.800\,\mu\mathrm m}{10.4\,\mu\mathrm m} \simeq2.8\times10^2\,\mathrm K. \]

Both calculations identify the relevant thermal scale; a thermal spectrum has no sharp upper photon-energy cutoff.

B.2

The mean photon count is \(p\tau\), and the mean dark count is \(i_d\tau\). Their Poisson variances equal their respective means.

Assuming independent noise sources, variances add:

\[ \boxed{\sigma_t^2=\sigma_r^2+i_d\tau+p\tau.} \]

Therefore,

\[ \boxed{\sigma_t=\sqrt{\sigma_r^2+i_d\tau+p\tau}.} \]

Subtracting the known mean dark count removes its mean contribution but leaves its statistical fluctuations.

B.3

The straight line on the graph corresponds approximately to

\[ i_d(T)=10^{\,14-100\,\mathrm K/T}\,\mathrm{s^{-1}}. \]

At \(T_p=7.5\,\mathrm K\),

\[ i_d=10^{14-100/7.5} =10^{2/3} \simeq4.64\,\mathrm{s^{-1}}. \]

Write \(N=p\tau\) for the expected photon count. The limiting condition is

\[ N=10\sqrt{\sigma_r^2+i_d\tau+N}. \]

Squaring and solving the resulting quadratic gives the positive root

\[ N=50+\sqrt{2500+100(\sigma_r^2+i_d\tau)}. \]

With \(\sigma_r=14\) and \(\tau=10^4\,\mathrm s\),

\[ N\simeq2.21\times10^3. \]

Thus

\[ \boxed{p_{\min}=\frac{N}{\tau} \simeq0.221\,\mathrm{s^{-1}} \approx0.22\,\mathrm{s^{-1}}.} \]

B.4

Each threshold photon has energy \(|\Delta E_g|\). With unit detection efficiency and lossless collection, the required power at the mirror is \(p|\Delta E_g|\). Hence

\[ I=\frac{p|\Delta E_g|}{A_{\mathrm{mirror}}}. \]

Substitution gives

\[ I=\frac{(0.221)(1.91\times10^{-20})}{25} \simeq1.69\times10^{-22}\,\mathrm{W/m^2}. \]

Therefore,

\[ \boxed{I\simeq1.7\times10^{-22}\,\mathrm{W/m^2}.} \]

C.1

Let \(A=A_{\mathrm{sheet}}\), and define the power emitted by one face of sheet \(i\) as

\[ q_i=\epsilon\sigma A T_i^4. \]

By Kirchhoff’s law, the constant emissivity also gives the absorptivity, so sheet 1 absorbs solar power

\[ P_\odot=\epsilon I_0A. \]

Following the stipulated left-to-right leakage model, define the net power received by sheet \(i+1\) from the preceding gap as

\[ F_i=\alpha(q_i-q_{i+1}). \]

The accompanying power escaping through that gap is

\[ L_i=\beta(q_i-q_{i+1}) =\frac{\beta}{\alpha}F_i. \]

Thus the total power entering a gap from its hotter side is

\[ F_i+L_i=rF_i, \qquad r=1+\frac{\beta}{\alpha}. \]

For each intermediate sheet, equilibrium requires

\[ F_{i-1}=rF_i, \qquad i=2,3,4. \]

The final sheet radiates \(q_5\) from its exposed face, so

\[ F_4=q_5. \]

Working back through the gaps,

\[ F_3=rq_5,\qquad F_2=r^2q_5,\qquad F_1=r^3q_5. \]

The temperature differences satisfy

\[ q_i-q_{i+1}=\frac{F_i}{\alpha}. \]

Adding all four differences gives

\[ q_1-q_5 =\frac{q_5}{\alpha}(1+r+r^2+r^3). \]

Define

\[ K=1+\frac{1+r+r^2+r^3}{\alpha}; \]

then \(q_1=Kq_5\).

Finally, the first sheet loses \(q_1\) through its Sun-facing surface and \(rF_1\) into the first gap:

\[ \epsilon I_0A=q_1+rF_1 =(K+r^4)q_5. \]

Using \(q_5=\epsilon\sigma AT_5^4\), we obtain

\[ \boxed{ T_5=\left[\frac{I_0}{\sigma(K+r^4)}\right]^{1/4} } \]

and, since \(T_1^4=KT_5^4\),

\[ \boxed{ T_1=\left[\frac{I_0}{\sigma}\frac{K}{K+r^4}\right]^{1/4}. } \]

The explicit factors of \(\epsilon A\) cancel; the emissivity still influences \(\alpha\) and \(\beta\).

C.2

The side length of a sheet is

\[ a=\sqrt{A}=\sqrt{200}=14.14\,\mathrm m. \]

The perimeter opening has area

\[ A_{\mathrm{gap}}=4ah =4(14.14)(0.25) =14.14\,\mathrm{m^2}. \]

Because the sheets are highly reflective and reflections are diffuse, approximate the radiation inside the box as well mixed after repeated reflections. Absorption probabilities are then proportional to absorbing area.

Each sheet has effective absorbing area \(\epsilon A\), while the opening acts as a black absorber with effective area \(A_{\mathrm{gap}}\). The total effective absorbing area is therefore

\[ 2\epsilon A+A_{\mathrm{gap}}. \]

The fraction ultimately absorbed by the other sheet is approximately

\[ \boxed{ \alpha\simeq \frac{\epsilon A}{2\epsilon A+A_{\mathrm{gap}}}. } \]

The fraction escaping is approximately

\[ \boxed{ \beta\simeq \frac{A_{\mathrm{gap}}}{2\epsilon A+A_{\mathrm{gap}}}. } \]

For \(\epsilon A=0.05(200)=10\,\mathrm{m^2}\),

\[ \boxed{\alpha\simeq\frac{10}{20+14.14}=0.293,} \]

\[ \boxed{\beta\simeq\frac{14.14}{20+14.14}=0.414.} \]

The remaining fraction, approximately \(\alpha\), returns to and is absorbed by the emitting sheet; accordingly, \(2\alpha+\beta=1\) in this estimate.

C.3

The preceding estimates give

\[ r=1+\frac{\beta}{\alpha}=2.414, \]

\[ r^4\simeq33.97,\qquad K\simeq80.60, \qquad K+r^4\simeq114.57. \]

Therefore,

\[ T_5= \left[ \frac{1360} {(5.670374419\times10^{-8})(114.57)} \right]^{1/4} \simeq120.3\,\mathrm K, \]

and

\[ T_1=K^{1/4}T_5\simeq360.4\,\mathrm K. \]

Thus

\[ \boxed{T_1\simeq360\,\mathrm K,\qquad T_5\simeq120\,\mathrm K.} \]

D.1

The required entries are:

Quantity Comparison
Internal energy \(U_1\ ?\ U_2\)
Temperature \(T_1\ ?\ T_2\)
Entropy \(S_1<S_2\)
Pressure \(P_1>P_2\)
Volume \(V_1<V_2\)

A pressure drop drives the flow through the resistive plug, so \(P_1>P_2\). Viscous dissipation makes the adiabatic process irreversible, giving \(S_2>S_1\).

Using the equal pipe cross sections shown, steady mass conservation gives

\[ \rho_1w_1=\rho_2w_2, \]

where \(w\) is bulk speed. Since \(w_2>w_1\), the density decreases and the volume occupied by one mole increases.

For a real gas, neither internal energy nor temperature has a universal direction of change during throttling. Their comparisons require further information about the states.

D.2

Follow one mole through the plug. The upstream gas does work \(P_1V_1\) on this mole, while it does work \(P_2V_2\) on the downstream gas.

There is no heat transfer. Neglecting the small change in bulk kinetic energy, the first law gives

\[ U_2-U_1=P_1V_1-P_2V_2. \]

Rearranging,

\[ \boxed{U_1+P_1V_1=U_2+P_2V_2.} \]

Thus the conserved quantity is the enthalpy,

\[ \boxed{H=U+PV.} \]

More precisely, steady adiabatic flow conserves \(h+w^2/2\) per unit mass. The stated approximation permits neglecting the kinetic-energy difference.

D.3

For clarity, write \(u\) and \(v\) for the specific internal energy and specific volume labelled \(U\) and \(V\) on the graphs.

The thermodynamic identity is

\[ du=T\,ds-P\,dv. \]

Along a constant-entropy curve,

\[ \left(\frac{\partial u}{\partial v}\right)_s=-P. \]

Therefore, the tangent to an isentrope at \((v,u)\) has equation

\[ u_{\mathrm{tan}}(x)=u-P(x-v) =(u+Pv)-Px. \]

Its intercept on the vertical axis is precisely the specific enthalpy:

\[ \boxed{h=u+Pv.} \]

The required construction on image 9 is:

  1. Locate \(v_2=0.100\,\mathrm{m^3/kg}\).
  2. Interpolate to the \(7.5\,\mathrm K\) isotherm.
  3. At this state, draw the tangent to the constant-entropy curve passing through it.
  4. Extend that tangent to \(v=0\); read its vertical intercept.

The state has approximately

\[ u_2\simeq26\,\mathrm{kJ/kg}. \]

The tangent slope is approximately

\[ -P_2\simeq-140\,\mathrm{kPa}. \]

Consequently,

\[ h_2=u_2+P_2v_2 \simeq26+(140)(0.100) =40\,\mathrm{kJ/kg}. \]

Thus

\[ \boxed{h_1=h_2\simeq4.0\times10^4\,\mathrm{J/kg}.} \]

For helium with molar mass approximately \(4.0\times10^{-3}\,\mathrm{kg/mol}\), this corresponds to

\[ H_{\mathrm{molar}}\simeq160\,\mathrm{J/mol}. \]

D.4

All admissible inlet states must have the same specific enthalpy \(h_0\simeq40\,\mathrm{kJ/kg}\).

The graphical construction on image 10 is:

  1. Mark the point \((0,h_0)\) on the vertical axis.
  2. Draw tangents from this point to several solid constant-entropy curves.
  3. Mark their points of tangency. Every such point has \(u+Pv=h_0\), so these points trace the constant-enthalpy curve.
  4. Find the highest-temperature dashed isotherm reached by this curve. At the maximum temperature, the constant-enthalpy curve touches that isotherm.

The limiting contact is close to the \(11\,\mathrm K\) isotherm, with graphical readings approximately

\[ v_1\sim1.0\times10^{-2}\,\mathrm{m^3/kg}, \qquad u_1\sim2.1\times10^4\,\mathrm{J/kg}. \]

The graph supports a maximum of roughly \(10.5\)–\(11\,\mathrm K\), so

\[ \boxed{T_{1,\max}\simeq11\,\mathrm K.} \]

D.5

The pressure is the magnitude of the isentropic tangent slope at the selected inlet state. Equivalently,

\[ P_1=\frac{h_0-u_1}{v_1}. \]

Using the approximate graphical coordinates,

\[ P_1\simeq \frac{(40-21)\times10^3}{1.0\times10^{-2}} =1.9\times10^6\,\mathrm{Pa}. \]

To the precision justified by the graph,

\[ \boxed{P_1\simeq2\times10^6\,\mathrm{Pa}.} \]

The solution was checked against the official solution.