Conside a cubical vessel of edge a having a small hole in one of its walls is r. At time `t=0` , it contains air at atmospheric pressure `p_(a) and temperature `T_(0)` . The temperature of the surrouding air is `T_(a)(gtT_(0)` . Find the amount of the gas (in moles) in the vessel at time t. Take `C_(v)` of air to be 5 R//2`.

Explanation

As the gas can leak out of the hole, the pressure inside the vessel will be equal to the atmospheric pressure `P_(a)` . Let n be the amount of the gas (moles) in the vessel at time t. Suppose an amount DeltaQ of heat is given to the gas in time dt. Its temperature increases by dT where
`DeltaQ=nC_(p)dT` . If the temperature of the gas is T at time t. we have `(DeltaQ)/(dt)=(T_(a)-T)/(r)` . or, `(c_(p)r)ndT=(T_(A)-(T)dt` . We have, `p_(a)a^(3)=nRT` . or, `ndt+Tdn=0` . or, `ndT=-Tdn` . Also, `T=p_(a)a^(3)/(nR)` . Using (ii) and (iii) in (i), `-C_(p)rP_(a)a^(3)/(nR)dn=(T_(a)-(p_(a)a^(3))/(nR))dt` . or, `(dn)/(nR(T_(a)-(p_(a)a^(3))/(nR)))=-(dt)/(C_(p)rp_(a)a^(3))` . or, `int_(on)^(n)(dn)/(nRT_(a)-p_(a)a^(3))=-int_(0)^(t)dt/(C_(p)rp_(a)a^(3))` . Where `n_(o)=(p_(a)a^(3))/(RT_(0))` . is the initial amount of the gas in the vessel . Thus, `(1)/(RT_(a))ln`(nRt_(a)-p_(a)a^(3))/(n_(0)RT_(a)-p_(a)a^(3))=-(t)/(C_(p)rp_(a)a^(3))` . or, `nRT_(a)-p_(a)a^(3)=(n_(0)RT_(a)-p_(a)a^(3))e-`^(RT_(a))/(c_(p)rp_(a)a^(3))t` . Writing `n_(0)=(p_(a)a^(3)/(RT_(0)` and C_(p)=C_(v)+R=(7R)/(2)` , `n=(p_(a)a^(3))/(RT_(a))[1+(T_(a)/(T_(0))-1)e-`^(2T_(a))/(7rp_(a)a^(3))t]` .