Новое решение
en/5.6.12.md
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| + | ### Statement | ||
| + | |||
| + | $5.6.12.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | 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\ | ||
| + | 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:\ | ||
| + | Let's go with a)\ | ||
| + | $W=\vec{F}\bullet\Delta l\vect{l}$\ | ||
| + | $W=F*Delta l*cos(0)$\ | ||
| + | And the cosine of 0 is 1\ | ||
| + | $W=F*Delta l$ and $F=P*S$\ | ||
| + | $W=P*S*Delta l$\ | ||
| + | The initial volume of the gas is:\ | ||
| + | $V=S*l$\ | ||
| + | And the final volume:\ | ||
| + | $2V=S*l'$\ | ||
| + | $2V-V=S*(l'-l)$\ | ||
| + | Then $V=S*Delta l$\ | ||
| + | Substituting this in the work we'll have:\ | ||
| + | $W=P*V$\ | ||
| + | For b)\ | ||
| + | We know that for a linear function\ | ||
| + | $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\ | ||
| + | $m=\frac{2P-P}{2V-V}$\ | ||
| + | $m=\frac{P}{V}$\ | ||
| + | And substituting this in $P=m*V + n$, we get\ | ||
| + | $P=P+n$, 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\ | ||
| + | $W=\frac{(2P+P)*(2V-V)}{2}$\ | ||
| + | So $W=\frac{3PV}{2}$\ | ||
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| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $5.6.12.$ [Insert the problem statement] | |||
| ### Solution | |||
| 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\ | |||
| 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:\ | |||
| Let's go with a)\ | |||
| $W=\vec{F}\bullet\Delta l\vect{l}$\ | |||
| $W=F*Delta l*cos(0)$\ | |||
| And the cosine of 0 is 1\ | |||
| $W=F*Delta l$ and $F=P*S$\ | |||
| $W=P*S*Delta l$\ | |||
| The initial volume of the gas is:\ | |||
| $V=S*l$\ | |||
| And the final volume:\ | |||
| $2V=S*l'$\ | |||
| $2V-V=S*(l'-l)$\ | |||
| Then $V=S*Delta l$\ | |||
| Substituting this in the work we'll have:\ | |||
| $W=P*V$\ | |||
| For b)\ | |||
| We know that for a linear function\ | |||
| $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\ | |||
| $m=\frac{2P-P}{2V-V}$\ | |||
| $m=\frac{P}{V}$\ | |||
| And substituting this in $P=m*V + n$, we get\ | |||
| $P=P+n$, 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\ | |||
| $W=\frac{(2P+P)*(2V-V)}{2}$\ | |||
| So $W=\frac{3PV}{2}$\ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||