| ### Statement | | ### Statement |
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| $5.6.26^*.$ A horizontally placed cylinder, containing $1$ mole of gas at an initial temperature $T_{0}$ and pressure $P_{0}$, is closed by a piston of cross-section $S$. To the right of the piston is a constant atmospheric pressure $P_{0}$. The gas is heated by a coil. When the piston moves, a friction force $F$ acts on it from the cylinder walls. Half of the heat generated by the friction of the piston against the cylinder walls goes into the gas. The internal energy of the gas is $U = cT$. How does the temperature of the gas depend on the amount of heat transferred to the gas by the coil? Plot the graph of this dependence. | | $5.6.26^*.$ A horizontally placed cylinder, containing $1$ mole of gas at an initial temperature $T_{0}$ and pressure $P_{0}$, is closed by a piston of cross-section $S$. To the right of the piston is a constant atmospheric pressure $P_{0}$. The gas is heated by a coil. When the piston moves, a friction force $F$ acts on it from the cylinder walls. Half of the heat generated by the friction of the piston against the cylinder walls goes into the gas. The internal energy of the gas is $U = cT$. How does the temperature of the gas depend on the amount of heat transferred to the gas by the coil? Plot the graph of this dependence. |
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| ### Solution | | ### Solution |
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| As the gas is heated, the pressure will first increase at a constant volume, since the friction force must be overcome to move the piston. Thus, the dependence of temperature on $Q$ will be different depending on whether the piston has started moving or not. We will assume that the dependence has one form as long as $Q \\le Q_1$, and another form when $Q \\ge Q_1$. | | As the gas is heated, the pressure will first increase at a constant volume, since the friction force must be overcome to move the piston. Thus, the dependence of temperature on $Q$ will be different depending on whether the piston has started moving or not. We will assume that the dependence has one form as long as $Q \\le Q_1$, and another form when $Q \\ge Q_1$. |
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| #### First segment | | #### First segment |
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| Let us consider the forces acting on the piston: | | Let us consider the forces acting on the piston: |
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| $$ \\vec{F}_{gas} + \\vec{F}_{fr} + \\vec{F}_{atm} = 0 $$ | | $$ \\vec{F}_{gas} + \\vec{F}_{fr} + \\vec{F}_{atm} = 0 $$ |
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| This equality will also hold during the uniform motion of the piston. For the piston to start moving, the force of the gas pressure must overcome the atmospheric pressure force and the maximum static friction force. Then, projecting the forces onto the horizontal axis, we obtain: | | This equality will also hold during the uniform motion of the piston. For the piston to start moving, the force of the gas pressure must overcome the atmospheric pressure force and the maximum static friction force. Then, projecting the forces onto the horizontal axis, we obtain: |
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| $$ F_{gas} = F_{fr} + F_{atm} $$ | | $$ F_{gas} = F_{fr} + F_{atm} $$ |
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| $$ pS = F + p_0 S $$ | | $$ pS = F + p_0 S $$ |
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| $$ p = p_0 + \\frac{F}{S} $$ | | $$ p = p_0 + \\frac{F}{S} $$ |
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| This is the pressure required for the piston to begin moving. | | This is the pressure required for the piston to begin moving. |
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| During the initial heating of the gas up to this pressure, the following holds true (since the process is isochoric): | | During the initial heating of the gas up to this pressure, the following holds true (since the process is isochoric): |
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| $$ \\frac{p_0}{T_0} = \\frac{p}{T} $$ | | $$ \\frac{p_0}{T_0} = \\frac{p}{T} $$ |
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| Then we can find the temperature at which the piston starts moving: | | Then we can find the temperature at which the piston starts moving: |
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| $$ T = T_0 \\frac{p}{p_0} = T_0 \\frac{p_0 + \\frac{F}{S}}{p_0} = T_0 \\left( 1 + \\frac{F}{p_0S}\\right) $$ | | $$ T = T_0 \\frac{p}{p_0} = T_0 \\frac{p_0 + \\frac{F}{S}}{p_0} = T_0 \\left( 1 + \\frac{F}{p_0S}\\right) $$ |
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| Since the process is initially isochoric, no work is done. According to the first law of thermodynamics: | | Since the process is initially isochoric, no work is done. According to the first law of thermodynamics: |
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| $$ Q = A' + \\Delta U = \\Delta U $$ | | $$ Q = A' + \\Delta U = \\Delta U $$ |
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| Knowing the dependence of the internal energy of the gas on temperature $U = cT$: | | Knowing the dependence of the internal energy of the gas on temperature $U = cT$: |
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| $$ Q = \\Delta U = U_1 - U_0 = cT - cT_0 $$ | | $$ Q = \\Delta U = U_1 - U_0 = cT - cT_0 $$ |
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| Finally: | | Finally: |
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| $$ T = T_0 + \\frac{Q}{c}, \\quad Q \\le Q_1 $$ | | $$ T = T_0 + \\frac{Q}{c}, \\quad Q \\le Q_1 $$ |
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| Here $Q_1$ is the amount of heat transferred to the gas before the piston starts moving: | | Here $Q_1$ is the amount of heat transferred to the gas before the piston starts moving: |
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| $$ Q_1 = c(T - T_0) = c\\left(T_0 \\left( 1 + \\frac{F}{p_0S}\\right) - T_0\\right) = \\frac{cT_0F}{p_0S} $$ | | $$ Q_1 = c(T - T_0) = c\\left(T_0 \\left( 1 + \\frac{F}{p_0S}\\right) - T_0\\right) = \\frac{cT_0F}{p_0S} $$ |
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| #### Second segment | | #### Second segment |
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| In this segment, the piston is moving. The friction force during the motion is constant, and the atmospheric pressure force is also constant, which means the gas pressure force is constant as well. Therefore, the process is isobaric, and the pressure in it is: | | In this segment, the piston is moving. The friction force during the motion is constant, and the atmospheric pressure force is also constant, which means the gas pressure force is constant as well. Therefore, the process is isobaric, and the pressure in it is: |
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| $$ p = p_0 + \\frac{F}{S} $$ | | $$ p = p_0 + \\frac{F}{S} $$ |
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| The amount of heat transferred to the gas from the coil consists of two parts — the heat transferred before the piston started moving $Q_1$ and the heat transferred after that $Q_2$: | | The amount of heat transferred to the gas from the coil consists of two parts — the heat transferred before the piston started moving $Q_1$ and the heat transferred after that $Q_2$: |
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| $$ Q = Q_1 + Q_2 $$ | | $$ Q = Q_1 + Q_2 $$ |
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| It is known that half of the heat released during the friction of the piston against the walls returns to the gas, so the first law of thermodynamics will look like this: | | It is known that half of the heat released during the friction of the piston against the walls returns to the gas, so the first law of thermodynamics will look like this: |
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| $$ Q_2 + Q_{fr} = A' + \\Delta U $$ | | $$ Q_2 + Q_{fr} = A' + \\Delta U $$ |
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| Let us consider the work of friction over the piston displacement $x$: | | Let us consider the work of friction over the piston displacement $x$: |
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| $$ A_{fr} = F x $$ | | $$ A_{fr} = F x $$ |
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| The change in the volume of the gas over this displacement is: | | The change in the volume of the gas over this displacement is: |
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| $$ \\Delta V = xS $$ | | $$ \\Delta V = xS $$ |
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| On the other hand, from the ideal gas law: | | On the other hand, from the ideal gas law: |
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| $$ \\Delta V = \\frac{\\nu R \\Delta T}{p} = \\frac{\\nu R \\Delta T}{p_0 + \\frac{F}{S}} $$ | | $$ \\Delta V = \\frac{\\nu R \\Delta T}{p} = \\frac{\\nu R \\Delta T}{p_0 + \\frac{F}{S}} $$ |
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| Then the displacement of the piston can be expressed as follows: | | Then the displacement of the piston can be expressed as follows: |
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| $$ x = \\frac{\\Delta V}{S} = \\frac{1}{S} \\frac{\\nu R \\Delta T}{p_0 + \\frac{F}{S}} = \\frac{\\nu R \\Delta T}{p_0S + F} $$ | | $$ x = \\frac{\\Delta V}{S} = \\frac{1}{S} \\frac{\\nu R \\Delta T}{p_0 + \\frac{F}{S}} = \\frac{\\nu R \\Delta T}{p_0S + F} $$ |
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| And the work of friction is: | | And the work of friction is: |
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| $$ A_{fr} = F \\frac{\\nu R \\Delta T}{p_0S + F} $$ | | $$ A_{fr} = F \\frac{\\nu R \\Delta T}{p_0S + F} $$ |
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| Half of this energy returns to the gas: | | Half of this energy returns to the gas: |
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| $$ Q_{fr} = \\frac{F \\nu R \\Delta T}{2(p_0 S + F)} $$ | | $$ Q_{fr} = \\frac{F \\nu R \\Delta T}{2(p_0 S + F)} $$ |
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| The work of the gas in the isobaric process is: | | The work of the gas in the isobaric process is: |
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| $$ A' = p \\Delta V = \\nu R \\Delta T $$ | | $$ A' = p \\Delta V = \\nu R \\Delta T $$ |
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| The change in the internal energy of the gas, as shown above: | | The change in the internal energy of the gas, as shown above: |
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| $$ \\Delta U = c \\Delta T $$ | | $$ \\Delta U = c \\Delta T $$ |
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| Let us write the first law of thermodynamics, substituting the obtained expressions for the heat returned from friction work, the work of the gas, and the change in the internal energy of the gas: | | Let us write the first law of thermodynamics, substituting the obtained expressions for the heat returned from friction work, the work of the gas, and the change in the internal energy of the gas: |
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| $$ Q_2 + \\frac{F \\nu R \\Delta T}{2(p_0 S + F)} = \\nu R \\Delta T + c \\Delta T $$ | | $$ Q_2 + \\frac{F \\nu R \\Delta T}{2(p_0 S + F)} = \\nu R \\Delta T + c \\Delta T $$ |
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| Rearranging the terms: | | Rearranging the terms: |
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| $$ Q_2 = \\Delta T \\left( \\nu R + c - \\frac{F \\nu R}{2(p_0 S + F)} \\right) $$ | | $$ Q_2 = \\Delta T \\left( \\nu R + c - \\frac{F \\nu R}{2(p_0 S + F)} \\right) $$ |
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| The change in temperature is calculated from the moment the piston starts moving: | | The change in temperature is calculated from the moment the piston starts moving: |
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| $$ \\Delta T = T - T_1 = T - T_0 \\left( 1 + \\frac{F}{p_0S}\\right) $$ | | $$ \\Delta T = T - T_1 = T - T_0 \\left( 1 + \\frac{F}{p_0S}\\right) $$ |
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| The amount of heat transferred to the gas from the coil after the piston starts moving: | | The amount of heat transferred to the gas from the coil after the piston starts moving: |
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| $$ Q_2 = Q - Q_1 = Q - \\frac{cT_0F}{p_0S} $$ | | $$ Q_2 = Q - Q_1 = Q - \\frac{cT_0F}{p_0S} $$ |
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| We obtain: | | We obtain: |
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| $$ Q - \\frac{cT_0F}{p_0S} = \\left(T - T_0 \\left( 1 + \\frac{F}{p_0S}\\right) \\right) \\left( \\nu R + c - \\frac{F \\nu R}{2(p_0 S + F)} \\right) $$ | | $$ Q - \\frac{cT_0F}{p_0S} = \\left(T - T_0 \\left( 1 + \\frac{F}{p_0S}\\right) \\right) \\left( \\nu R + c - \\frac{F \\nu R}{2(p_0 S + F)} \\right) $$ |
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| $$ Q - \\frac{cT_0F}{p_0S} = \\left(T - T_0 - \\frac{T_0 F}{p_0S} \\right) \\left( \\nu R + c - \\frac{F \\nu R}{2(p_0 S + F)} \\right) $$ | | $$ Q - \\frac{cT_0F}{p_0S} = \\left(T - T_0 - \\frac{T_0 F}{p_0S} \\right) \\left( \\nu R + c - \\frac{F \\nu R}{2(p_0 S + F)} \\right) $$ |
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| Expanding the brackets: | | Expanding the brackets: |
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| $$ Q - \\frac{cT_0F}{p_0S} = \\nu R T - \\nu R T_0 - \\frac{\\nu R T_0 F}{p_0S} + cT - cT_0 - \\frac{cT_0F}{p_0S} - \\frac{T F \\nu R}{2(p_0 S + F)} + \\frac{T_0 F \\nu R}{2(p_0 S + F)} + \\frac{T_0 F^2 \\nu R}{2p_0S(p_0 S + F)} $$ | | $$ Q - \\frac{cT_0F}{p_0S} = \\nu R T - \\nu R T_0 - \\frac{\\nu R T_0 F}{p_0S} + cT - cT_0 - \\frac{cT_0F}{p_0S} - \\frac{T F \\nu R}{2(p_0 S + F)} + \\frac{T_0 F \\nu R}{2(p_0 S + F)} + \\frac{T_0 F^2 \\nu R}{2p_0S(p_0 S + F)} $$ |
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| Canceling like terms: | | Canceling like terms: |
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| $$ Q = \\nu RT - \\nu RT_0 - \\frac{\\nu RT_0F}{p_0S} + cT - cT_0 - \\frac{TF\\nu R}{2(p_0S+F)} + \\frac{T_0F\\nu R}{2(p_0S+F)} + \\frac{T_0F^2\\nu R}{2p_0S(p_0S+F)} $$ | | $$ Q = \\nu RT - \\nu RT_0 - \\frac{\\nu RT_0F}{p_0S} + cT - cT_0 - \\frac{TF\\nu R}{2(p_0S+F)} + \\frac{T_0F\\nu R}{2(p_0S+F)} + \\frac{T_0F^2\\nu R}{2p_0S(p_0S+F)} $$ |
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| Grouping the coefficients for $T$ and $T_0$: | | Grouping the coefficients for $T$ and $T_0$: |
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| $$ Q = T \\left(\\nu R + c - \\frac{F\\nu R}{2(p_0S+F)}\\right) - T_0\\left( \\nu R + c + \\frac{\\nu RF}{p_0S} - \\frac{F\\nu R}{2(p_0S+F)} - \\frac{F^2\\nu R}{2p_0S(p_0S+F)}\\right) $$ | | $$ Q = T \\left(\\nu R + c - \\frac{F\\nu R}{2(p_0S+F)}\\right) - T_0\\left( \\nu R + c + \\frac{\\nu RF}{p_0S} - \\frac{F\\nu R}{2(p_0S+F)} - \\frac{F^2\\nu R}{2p_0S(p_0S+F)}\\right) $$ |
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| Transforming the last two terms in the last bracket: | | Transforming the last two terms in the last bracket: |
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| $$ - \\frac{F\\nu R}{2(p_0S+F)} - \\frac{F^2\\nu R}{2p_0S(p_0S+F)} = - \\frac{F\\nu R}{2(p_0S+F)}\\left(1 + \\frac{F}{p_0S}\\right) = $$ | | $$ - \\frac{F\\nu R}{2(p_0S+F)} - \\frac{F^2\\nu R}{2p_0S(p_0S+F)} = - \\frac{F\\nu R}{2(p_0S+F)}\\left(1 + \\frac{F}{p_0S}\\right) = $$ |
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| $$ = - \\frac{F\\nu R}{2(p_0S+F)} \\cdot \\frac{p_0S+F}{p_0S} = - \\frac{F\\nu R}{2p_0S} $$ | | $$ = - \\frac{F\\nu R}{2(p_0S+F)} \\cdot \\frac{p_0S+F}{p_0S} = - \\frac{F\\nu R}{2p_0S} $$ |
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| Then the entire bracket becomes: | | Then the entire bracket becomes: |
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| $$ \\nu R + c + \\frac{\\nu RF}{p_0S} - \\frac{F\\nu R}{2p_0S} = \\nu R + c + \\frac{\\nu RF}{2p_0S} $$ | | $$ \\nu R + c + \\frac{\\nu RF}{p_0S} - \\frac{F\\nu R}{2p_0S} = \\nu R + c + \\frac{\\nu RF}{2p_0S} $$ |
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| And the expression now looks like this: | | And the expression now looks like this: |
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| $$ Q = T \\left(\\nu R + c - \\frac{F\\nu R}{2(p_0S+F)}\\right) - T_0\\left( \\nu R + c + \\frac{\\nu RF}{2p_0S}\\right) $$ | | $$ Q = T \\left(\\nu R + c - \\frac{F\\nu R}{2(p_0S+F)}\\right) - T_0\\left( \\nu R + c + \\frac{\\nu RF}{2p_0S}\\right) $$ |
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| Then, expressing $T$: | | Then, expressing $T$: |
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| $$ T = \\frac{Q + T_0\\left(\\nu R + c + \\frac{\\nu RF}{2p_0S}\\right)}{\\nu R + c - \\frac{F\\nu R}{2(p_0S+F)}} $$ | | $$ T = \\frac{Q + T_0\\left(\\nu R + c + \\frac{\\nu RF}{2p_0S}\\right)}{\\nu R + c - \\frac{F\\nu R}{2(p_0S+F)}} $$ |
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| Substituting $\\nu = 1$, we get the answer for $Q \\ge Q_1$: | | Substituting $\\nu = 1$, we get the answer for $Q \\ge Q_1$: |
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| $$ T = \\frac{Q + T_0\\left(R + c + \\frac{RF}{2p_0S}\\right)}{R + c - \\frac{FR}{2(p_0S+F)}} $$ | | $$ T = \\frac{Q + T_0\\left(R + c + \\frac{RF}{2p_0S}\\right)}{R + c - \\frac{FR}{2(p_0S+F)}} $$ |
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| #### Answer | | #### Answer |
