At \(t=0\) \(50\) \(0\) \(0\)
At \(t=50\,min\) \(50-p_1\) \(2p_1\) \(\frac {p_1}{2}\)
Total pressure at \(50\,\min\)
\(=\,50-p_1+2p_1+\frac {p_1}{2}=87.5\)
\(50+\frac {3p_1}{2}=87.5\)
\(\frac{{3{p_1}}}{2}\, = \,37.5\)
\(\therefore \,\,{p_1}\, = \frac{{\,37.5\, \times 2}}{3}\, = \,25\)
At \(t=100\,min\) \(50-p_2\) \(2p_2\) \(\frac {p_2}{2}\)
\(50\,\min\) is half life period
For \(100\,\min\) i.e., for \(2\) half lives \(50-p_2=12.5\)
\(\therefore \,\,{p_2}\, = 37.5\,mm\) of \(Hg\)
Total pressure at \(100\,\min\)
\( = {\mkern 1mu} 50 - {p_2} + 2{p_2} + \frac{{{p_2}}}{2}\)
\( = {\mkern 1mu} 50 + \frac{{3{p_2}}}{2}\, = \,50 + \frac{3}{2} \times 37.5\)
\(=\,50\,+\,56.25\)
\(=\,106.25\,mm\) of \(Hg\)
$T$ (in, $K$) $- 769$ , $1/T$ (in, $K^{-1}$ ) $- 1.3\times 10^{-3},$
$\log_{10}K - 2.9\,T$ (in, $K$) $- 667$, $1/T$ (in, $K^{-1}) - 1.5\times 10^{-3}$, $\log_{10}\,K - 1.1$