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Question-31326




Question Number 31326 by rahul@ last updated on 06/Mar/18
Commented by rahul@ last updated on 06/Mar/18
plz solve it
$${plz}\:{solve}\:{it} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 20/May/18
when time t pressure=p  volume=v  in time t+dt , volume=v−dv  pressure=p+dp  (v−dv)(p+dp)=pv  pv+vdp−pdv=pv  vdp=pdv  (dp/p)=(dv/v)  lnp=lnv+lnk  ln((p/v))=lnk  p=vk  (dp/dt)=(dv/dt)×k  (dp/dt)=rk  (dv/dt)=r   r should be negative because dv=−ve  dt=+ve    dv=−rdt  v=−rt+c  t=0  v=v_0   v=−rt+v_0   v_o /2=−rt+v_0   rt=(v_0 /2)  t=(v_0 /(2r))    pls check whether ans is right or wrong  pls comment
$${when}\:{time}\:{t}\:{pressure}={p} \\ $$$${volume}={v} \\ $$$${in}\:{time}\:{t}+{dt}\:,\:{volume}={v}−{dv}\:\:{pressure}={p}+{dp} \\ $$$$\left({v}−{dv}\right)\left({p}+{dp}\right)={pv} \\ $$$${pv}+{vdp}−{pdv}={pv} \\ $$$${vdp}={pdv} \\ $$$$\frac{{dp}}{{p}}=\frac{{dv}}{{v}} \\ $$$${lnp}={lnv}+{lnk} \\ $$$${ln}\left(\frac{{p}}{{v}}\right)={lnk} \\ $$$${p}={vk} \\ $$$$\frac{{dp}}{{dt}}=\frac{{dv}}{{dt}}×{k} \\ $$$$\frac{{dp}}{{dt}}={rk} \\ $$$$\frac{{dv}}{{dt}}={r}\:\:\:{r}\:{should}\:{be}\:{negative}\:{because}\:{dv}=−{ve} \\ $$$${dt}=+{ve} \\ $$$$ \\ $$$${dv}=−{rdt} \\ $$$${v}=−{rt}+{c} \\ $$$${t}=\mathrm{0} \\ $$$${v}={v}_{\mathrm{0}} \\ $$$${v}=−{rt}+{v}_{\mathrm{0}} \\ $$$${v}_{{o}} /\mathrm{2}=−{rt}+{v}_{\mathrm{0}} \\ $$$${rt}=\frac{{v}_{\mathrm{0}} }{\mathrm{2}} \\ $$$${t}=\frac{{v}_{\mathrm{0}} }{\mathrm{2}{r}} \\ $$$$ \\ $$$${pls}\:{check}\:{whether}\:{ans}\:{is}\:{right}\:{or}\:{wrong} \\ $$$${pls}\:{comment} \\ $$

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