Question

$500ml$ of a vessel contains an ideal gas saturated with water vapours at $3.2$ atm and $47^{\circ }C$. Now the entire mass is transferred into vessel of $1800ml$ capacity maintained at $87^{\circ }C$. The final pressure (in atm) becomes (vapour pressure of water at $47^{\circ }C$ and $87^{\circ }C$ are $0.2$ atm and $0.6$ atm, respectively)

Answer 1

Answer: 1

$V_1=500ml$
$V_2=1800ml$
$P_{\text{total }}=3.2atm$
$T_2=360K$
$T_1=320K$
$V.P.=0.6atm$
$V.P.=0.2atm$
$P_{\text{gas }}=3.2-0.2=3$ atm
$P_{gas}=3.2-0.2$
For $H_{2}O$ (vap.)
$\frac{0.2\times 500}{320}=\frac{P_2\times 1800}{360}$
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
$=\frac{3\times 500}{320}=\frac{P_2\times 1800}{360}$
$P_2=\frac{20}{320}=\frac{2}{32}atm(<V\cdot P.)$
$P_2=\frac{6\times 500}{320\times 10}=\frac{30}{320}atm$
$P_{\text{total }}=\frac{30}{32}+\frac{2}{32}=1atm$