Question

The temperature of a certain mass of gas is double if initial gas is at 1 atm pressure. Find the % increased in pressure.

Answer 1

we know that for an ideal gas
pv=nrt
and hence
v=nrt/p
in the question it is given that the same gas is kept afterwards at a lower pressure and hence the no. of moles is same
so we know that the value of n,r,t are same (r is gas constant)
so
v 1/p
we also knoe that the final volume of gas is 75% of initial volume
=850*75/100 ml
so
volume initially(L)/volume final(L) =pressure final(atm) /pressure innitially(atm)
so
(850/1000) / (850*75/(100*1000))= (760/760) /
so
=0.75 atm
which equals 570mm of hg