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| ### Statement |
| ### Statement |
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| $5.6.12.$ [Insert the problem statement] |
| $5.6.12.$ In a cylindrical vessel, a movable piston covers the volume of gas V at pressure |
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| P. There is a vacuum on the other side of the piston. The plunger is released. |
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| What work will the gas do on the piston, if the volume of gas when moving the |
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| piston will double, and its pressure will be: a) remain constant; b) increase |
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| with increasing volume linearly to a pressure of 2P? |
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| ### Solution |
| ### Solution |
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| We have a cylindrical vessel, with a movable piston with a gas on its left side and vacuum on the right side. The gas is at pressure $P$ and volume $V$. We need the work done by the gas on the piston, if the volume of the gas when moving the piston will double, and its pressure will be:\ |
| We have a cylindrical vessel, with a movable piston with a gas on its left side, and vacuum on the right side. The gas is at pressure P and volume V. We need the work done by the gas on the piston, if the volume of the gas when moving the piston will double, and its pressure will be:\ |
| a)remain constant\ |
| a)remain constant\ |
| b)increase with increasing volume linearly yo a pressure of $2P$\ |
| b)increase with increasing volume linearly yo a pressure of $2P$\ |
| Assuming that we're working with ideal gases, and gas pushes the piston with a constant force, the work will be:\ |
| Assuming that we're working with ideal gases, and gas pushes the piston with a constant force, the work will be:\ |
| Let's go with a)\ |
| Let's go with a)\ |
| $W=\vec{F}\bullet\Delta l\vect{l}$\ |
| $W=\vec{F}\bullet\vec{\Delta l}$\ |
| $W=F*Delta l*cos(0)$\ |
| $W=F\Delta lcos(0)$\ |
| And the cosine of 0 is 1\ |
| And the cosine of 0 is 1\ |
| $W=F*Delta l$ and $F=P*S$\ |
| $W=F\Delta l$ and $F=PS$\ |
| $W=P*S*Delta l$\ |
| Where S is the cross sectional area \ |
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| $W=PS\Delta l$\ |
| The initial volume of the gas is:\ |
| The initial volume of the gas is:\ |
| $V=S*l$\ |
| $V=Sl$\ |
| And the final volume:\ |
| And the final volume:\ |
| $2V=S*l'$\ |
| $2V=Sl'$\ |
| $2V-V=S*(l'-l)$\ |
| $2V-V=S(l'-l)$\ |
| Then $V=S*Delta l$\ |
| Then $V=S\Delta l$\ |
| Substituting this in the work we'll have:\ |
| Substituting this in the work we'll have:\ |
| $W=P*V$\ |
| $W=PV$\ |
| For b)\ |
| For b)\ |
| We know that for a linear function\ |
| We know that for a linear function\ |
| $m=\frac{y_2 -y_1}{x_2 -x_1}\ |
| $m=\frac{y_2 -y_1}{x_2 -x_1}$\ |
| In our case, we have a $P=f(V)$ function, the final pressure is $2P$ and the final volume $2V$, so\ |
| In our case, we have a $P=f(V)$ function, the final pressure is $2P$ and the final volume $2V$, so\ |
| $m=\frac{2P-P}{2V-V}$\ |
| $m=\frac{2P-P}{2V-V}$\ |
| $m=\frac{P}{V}$\ |
| $m=\frac{P}{V}$\ |
| And substituting this in $P=m*V + n$, we get\ |
| And substituting this in $P=mV + n$, we get\ |
| $P=P+n$, no $n=0$\ |
| $P=P+n$, no $n=0$\ |
| If we graph this function, the work will be the area of the trapezoid under this two points, wich is (bass minor + base major) times height over 2\ |
| If we graph this function, the work will be the area of the trapezoid under this two points, wich is (bass minor + base major) times height over 2\ |
| $W=\frac{(2P+P)*(2V-V)}{2}$\ |
| $W=\frac{(2P+P)(2V-V)}{2}$\ |
| So $W=\frac{3PV}{2}$\ |
| So $W=\frac{3PV}{2}$ |
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| #### Answer |
| #### Answer |
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| [Insert a concise answer or boxed result] |
| a) $W=PV$\ |
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| b) $W=\frac{3PV}{2}$ |
