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Question Number 32316 by NECx last updated on 23/Mar/18
A gas expands according to the law pv=constant,where p is the pressure and v is the volume of the gas.Initially,v=1000m^3 and p=40N/m^2 .If the pressure is decreased at the rate of 5Nm^(−2) min^(−1) find the rate at which the gas is expanding when its volume is 2000m^(3.)
$${A}\:{gas}\:{expands}\:{according}\:{to}\:{the} \\ $$$${law}\:{pv}={constant},{where}\:{p}\:{is}\:{the} \\ $$$${pressure}\:{and}\:{v}\:{is}\:{the}\:{volume}\:{of} \\ $$$${the}\:{gas}.{Initially},{v}=\mathrm{1000}{m}^{\mathrm{3}} \:{and} \\ $$$${p}=\mathrm{40}{N}/{m}^{\mathrm{2}} .{If}\:{the}\:{pressure}\:{is} \\ $$$${decreased}\:{at}\:{the}\:{rate}\:{of}\:\mathrm{5}{Nm}^{−\mathrm{2}} {min}^{−\mathrm{1}} \\ $$$${find}\:{the}\:{rate}\:{at}\:{which}\:{the}\:{gas}\:{is} \\ $$$${expanding}\:{when}\:{its}\:{volume}\:{is} \\ $$$$\mathrm{2000}{m}^{\mathrm{3}.} \\ $$
Commented by NECx last updated on 23/Mar/18
I did it this way but the textbook says its wrong: pv=constant(p_1 v_1 =p_2 v_2 ) ∴ 1000×40=2000×p_2 p_2 =20Nm^(−2) ∴dp=20−40=−20Nm^(−2) ⇒(dp/dt)=5⇒dt=20/5=4min ∴dv/dt=(2000−1000)/4 =250m^3 min^(−1) please correct me where I got it wrong.
$${I}\:{did}\:{it}\:{this}\:{way}\:{but}\:{the}\:{textbook} \\ $$$${says}\:{its}\:{wrong}: \\ $$$$ \\ $$$${pv}={constant}\left({p}_{\mathrm{1}} {v}_{\mathrm{1}} ={p}_{\mathrm{2}} {v}_{\mathrm{2}} \right) \\ $$$$\therefore\:\mathrm{1000}×\mathrm{40}=\mathrm{2000}×{p}_{\mathrm{2}} \\ $$$$\:\:\:\:{p}_{\mathrm{2}} =\mathrm{20}{Nm}^{−\mathrm{2}} \\ $$$$\therefore{dp}=\mathrm{20}−\mathrm{40}=−\mathrm{20}{Nm}^{−\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\frac{{dp}}{{dt}}=\mathrm{5}\Rightarrow{dt}=\mathrm{20}/\mathrm{5}=\mathrm{4}{min} \\ $$$$ \\ $$$$\therefore{dv}/{dt}=\left(\mathrm{2000}−\mathrm{1000}\right)/\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{250}{m}^{\mathrm{3}} {min}^{−\mathrm{1}} \\ $$$$ \\ $$$${please}\:{correct}\:{me}\:{where}\:{I}\:{got}\:{it} \\ $$$${wrong}. \\ $$
Commented by NECx last updated on 23/Mar/18
textbook says 500m^3 min^(−1)
$${textbook}\:{says}\:\mathrm{500}{m}^{\mathrm{3}} {min}^{−\mathrm{1}} \\ $$
Commented by mrW2 last updated on 23/Mar/18
v=(c/p) (dv/dt)=−(c/p^2 )×(dp/dt)=−(v/p)×(dp/dt) at t=t_2 : v=2000 m^3 and p=20 N/m^2 ⇒(dv/dt)=−((2000)/(20))×(−5)=500 m^3 /min ×××××× p(t)=p_0 −5t v(t)=(c/(p_0 −5t)) v_0 =(c/p_0 ) ⇒v(t)=((p_0 v_0 )/(p_0 −5t))=(v_0 /(1−((5t)/p_0 ))) (dv/dt)≠constant
$${v}=\frac{{c}}{{p}} \\ $$$$\frac{{dv}}{{dt}}=−\frac{{c}}{{p}^{\mathrm{2}} }×\frac{{dp}}{{dt}}=−\frac{{v}}{{p}}×\frac{{dp}}{{dt}} \\ $$$${at}\:{t}={t}_{\mathrm{2}} : \\ $$$${v}=\mathrm{2000}\:{m}^{\mathrm{3}} \:{and}\:{p}=\mathrm{20}\:{N}/{m}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{dv}}{{dt}}=−\frac{\mathrm{2000}}{\mathrm{20}}×\left(−\mathrm{5}\right)=\mathrm{500}\:{m}^{\mathrm{3}} /{min} \\ $$$$ \\ $$$$×××××× \\ $$$${p}\left({t}\right)={p}_{\mathrm{0}} −\mathrm{5}{t} \\ $$$${v}\left({t}\right)=\frac{{c}}{{p}_{\mathrm{0}} −\mathrm{5}{t}} \\ $$$${v}_{\mathrm{0}} =\frac{{c}}{{p}_{\mathrm{0}} } \\ $$$$\Rightarrow{v}\left({t}\right)=\frac{{p}_{\mathrm{0}} {v}_{\mathrm{0}} }{{p}_{\mathrm{0}} −\mathrm{5}{t}}=\frac{{v}_{\mathrm{0}} }{\mathrm{1}−\frac{\mathrm{5}{t}}{{p}_{\mathrm{0}} }} \\ $$$$\frac{{dv}}{{dt}}\neq{constant} \\ $$
Commented by NECx last updated on 23/Mar/18
thank you so much
$${thank}\:{you}\:{so}\:{much} \\ $$

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