Two cylinders, A and B, have equal inside diameters and pistons of negligible mass connect...

SKETCH:

RESEARCH:

(a) For cylinder A changes are adiabatic \((\mathrm{Q}=0)\) and so (by the First Law Thermodynamics) \(\Delta E_{=i}=-W\). For adiabatic process:

\(p V_{Y}=\) constant

or equivalently

\(T V-1=\) constant \(\quad(2)\)

Work, \(W\), done by the argon is at the expense of its internal energy. Work done during an adiabatic expansion as temperature changes from TAi to TAf is:

\(W=n_{A} R\left(T_{A f}-T_{A i}\right) /(1-g)\)

Argon is a monotonic gas. According to kinetic theory of gases considering the elastic collisions of molecules with the walls of the container (see Chapter 19 for details) it can be shown that for a monotonic gas: \(\mathrm{Y}=5 / 3\).

(b) For cylinder B changes occur isothermally so (by the First Law Thermodynamics) \(\Delta E_{z=1}=0\) and \(Q=W\). For a gas undergoing compression isothermally:

\(p V=\) constant.

Work done on the gas during an isothermal compression as volume changes from VBi to VBf is:

\(W=n_{\mathrm{s}} R T \ln \left(V_{\mathrm{J} /} / V_{\mathrm{J}}\right)\),

(5)

and this is expelled as heat to the thermostat to maintain the temperature.

(c) After the piston is released, the connected pistons move so that the argon expands, while the oxygen contracts until the pressures are equal in each cylinder:

\(p \mathrm{Af}=p \mathrm{Bf}\)

\(p_{\lambda f}=p_{A} V_{A}{ }^{3} / V_{M^{\prime}}{ }^{3}\)

For cylinder B using equation (4): \(p_{\bar{\Sigma}} V_{\mathrm{si}}=p_{\mathrm{si}} V_{\mathrm{JI}} \mathrm{so}\) that

SIMPLIFY:

(a)For cylinder A using equation (2): \(T V_{Y}-1=\) constant and \(y=5 / 3\) for argon. Therefore \(y-1=\) \(2 / 3\), and \(T V^{\gamma-1}=T V^{23}\), as required. For cylinder \(\mathrm{A}\) it is given that when the piston is released the

equivalently (since \(82 / 3=4)\) that

TAf \(=1 / 4\) TAi.

(b) For cylinder B it is given that the density, \(\rho B f\), of the oxygen after the compression is twice the original density, \(\rho\) Bi. Density is \(\rho=m / V\) and this means that the volume of oxygen in cylinder B after the compression is \(V_{\mathrm{sE}}=m / \gamma_{\mathrm{s} E}=m / 2 r_{\mathrm{s}} .\) That is,

\(V_{\mathrm{JE}}=1 / 2 V_{\overline{\mathbf{x}}}\)

CALCULATE:

(a) For cylinder A: Mass of Argon \(=0.320 \mathrm{~kg}=320 . \mathrm{g} .\) Number of moles of Argon \(\mathrm{nA}=\) mass in grams/molar mass \(=320 . / 39.95=8.0100\)

(b) Using equations (6), (7) and (8):

Now using result (10) and the given fact that for cylinder A the volume increases from VAi to VAf \(=8\) VAi implies that in equation \((11): \mathrm{pAf}=\mathrm{pAi}(1 / 8) 5 / 3=\mathrm{pBf}=2 \mathrm{pBi}=2 \mathrm{~atm}=202.65 \mathrm{kPa}\), since

it is given that the original pressure in cylinder \(\mathrm{B}\) is normal atmospheric pressure \((101.325 \mathrm{kPa})\). Therefore the original pressure of the argon in cylinder \(\mathrm{A}\) is \(\mathrm{p}_{k}=\mathrm{s}^{5 \mathrm{~s}} \cdot(2 \mathrm{~atm})=64 \mathrm{~atm}=6484.8 \mathrm{kPa}\). The final pressure after expansion is \(\mathrm{pAf}=2 \mathrm{~atm}=202.65 \mathrm{kPa}\).

(c) Work done during the adiabatic expansion using equations (3) and (9) is:

\(W=n A R(T A f-T A i) /(1-\gamma)=(8.0100\) mol \()(8.31 \mathrm{~J} \mathrm{~mol}-1 \mathrm{~K}-1)(1 / 4 \mathrm{TA}-\mathrm{TAi}) /(1-5 / 3)=74.79 \mathrm{TAi}\)

JK-1

This can be equated to the heat received by the thermostat \(7.479 \cdot 10^{4} \mathrm{~J}\). This means that for the argon: \(\quad \mathrm{TAi}=1000 . \mathrm{K} .\) Using equation \((9): \mathrm{TAf}=1 / 4 \mathrm{TAi}=250 . \mathrm{K}\).

(d) For an ideal gas \(\mathrm{pV}=\mathrm{nRT}\) and so for the argon, using the above results for TAi, TAf, pAi, pAf

\(V_{x}=n_{A} R T_{x} / p_{x}=(8.00 \mathrm{~mol})\left(8.31 \mathrm{~mol}^{-1} \mathrm{JK}^{-1}\right)(1000 . \mathrm{K}) /\left(64 \cdot 101.325 \cdot 10^{3} \mathrm{~Pa}\right)=0.01025 \mathrm{~m}^{3}\)

\(V_{A t}=n_{\mathrm{A}} R T_{A t} / p_{A f}=(8.00 \mathrm{~mol})\left(8.31 \mathrm{~mol}^{-1} \mathrm{JK}^{-1}\right)(250 . \mathrm{K}) /\left(2 \cdot 101.325 \cdot 10^{3} \mathrm{~Pa}\right)=0.08201 \mathrm{~m}^{3}\) (check:

\(\left.V_{A f}=0.08201 \mathrm{~m}^{3}=8 V_{x}=8 \cdot 0.01025 \mathrm{~m}^{3}\right)\)

(e) Since the total volumes of cylinders \(A\) and \(B\) must remain constant it follows that:

So the total volume at any time is:

\(0.01025+0.1435=0.1537 \mathrm{~m} 3\)

\((\) check: \(V A f+V B f=0.08201 m 3+0.1435 / 2 m 3=0.1537 \mathrm{~m} 3)\)

(f) For cylinder \(\mathrm{B}\) : It is in contact with a thermostat that maintains the cylinder at a constant temperature of \(270 \mathrm{C}=273+27=300 . \mathrm{K}\). Work done during an isothermal compression using equation (5) is:

\(W=n_{\mathrm{s}} R T \ln \left(V_{\mathrm{I}:} / V_{\mathrm{J}}\right)=n\left(8.31 \mathrm{~mol}^{-1} \mathrm{JK}^{-1}\right)(300 \cdot \mathrm{K}) \ln \left(1 / 2 V_{\mathrm{si}} / V_{\mathrm{s}}\right)=-n_{\overline{\mathrm{s}}}\left(1728 \mathrm{~J} \mathrm{~mol}^{-1}\right)\)

where \(n B\) is the number of moles of oxygen in cylinder \(B\).

(g) The oxygen, maintained at 300. \(\mathrm{K}\) by the thermostat, is warmer than the argon \((250 . \mathrm{K})\) after the pressures equal in the cylinders. When the valve is opened to connect the two cylinders, the equilibrium temperature will be \(300 . \mathrm{K}\) since the thermostat will maintain the now connected cylinders at this temperature. The final pressure in the connected cylinders can be found using \(p V=n R T\) with \(T=300 . \mathrm{K}\) and the known \(\mathrm{n}\), the total number of moles of oxygen and argon. Additionally, equation (13) can be used to find the number of moles of oxygen:

\(1728 n_{\mathrm{s}}=7.479 \cdot 10^{4} \mathrm{nB}=43.28\). The total number of moles of gas is: \(8+43.28=51.28\), so that the pressure in the connectedcylinders is:

\(p=n R T / V=(51.28 \mathrm{~mol})\left(8.31 \mathrm{~mol}^{-1} \mathrm{~J} \mathrm{~K}^{-1}\right)(300 . \mathrm{K}) /\left(0.15375 \mathrm{~m}^{3}\right)\)

\(=831.486 \mathrm{kPa}=8.206 \mathrm{~atm}\)

ROUND: Quote all values to three significant figures:

\(\mathrm{pAi}=64.0 \mathrm{~atm}=6.48 \mathrm{MPa}\)

\(\mathrm{pAf}=2.00 \mathrm{~atm}=0.0203 \mathrm{MPa}\)

\(\mathrm{TAi}=1.00 \cdot 10^{3} \mathrm{~K}\)

\(\mathrm{TAf}=250 . \mathrm{K}\)

\(\mathrm{VAi}=0.0103 \mathrm{~m} 3\)

\(V A f=0.0820 \mathrm{~m} 3\)

Final pressure when connected is: \(p=0.831 \mathrm{MPa}=8.21\) atm.

DOUBLE CHECK:The final pressure when pistons were connected, after the valve was opened is more than it was when they were not connected.  This is as expected since heat flows into the cylinders to heat the argon to 300 K. The final pressure is less than it initially was when the piston was fixed, again this seems reasonable as at that time the argon was much hotter.