A container contains hydrogen gas at pressure P and temperature T. Another identical container contains helium gas at pressure 2P and temperature `T//2`. The ratio of the number of molecules of the two gases is

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]` .