$(a)\;432.5K\qquad(b)\;312.3K\qquad(c)\;337.5K\qquad(d)\;234.3K$

Initially at lower end

P = 76cm of Hg + 76cm of air = 152cm

T = 300K

$V = \large\frac{V_1}{2}$

Where $V_1$ is the volume of the cylinder

Finally at lower end

P = 76cm of air + 38cm of Hg = 114 cm

T = ?

$V = \large\frac{3V_1}{4}$

$\therefore \large\frac{P_1V_1}{T_1} = \large\frac{P_2V_2}{T_2}$

$\large\frac{152\times V_1}{2\times300} = \large\frac{114\times3V_1}{4\times T}$

$\therefore T = \large\frac{114\times3\times2\times300}{152\times4}$

= 337.5K

Hence answer is (c)

Ask Question

Tag:MathPhyChemBioOther

Take Test

...