Question

A vessel of volume $50000c{m}^3$ contains an ideal gas at $0^{o}C$. After a portion is let out, the pressure in the vessel decreases by 1 atm in constant temperature conditions. If the density under normal conditions is $1.6kg{m}^{-3}$ then find the mass of the released gas in decagrams :

Answer 1

$PV=nRT$

$\Rightarrow P=\frac{\rho }{M_W}RT$

$\Rightarrow \rho =\frac{P{M}_W}{RT}$

Mass let out $=V{\rho }_o=V\rho$

$=V\left(\frac{P_0{M}_W}{RT}\right)-V\left(\frac{P{M}_W}{RT}\right)=\frac{V{M}_W}{RT}{P}_0-P)$

$=\frac{V{M}_W}{RT}\left(\Delta P\right)$

$=\frac{V_{{\rho }_o}\cdot \Delta P}{P_0}$

$=\frac{(50000\times {10}^{-6})\left(1.6\right)\left(1\right)}{\left(1\right)}kg=8\times {10}^{-2}kg=80g=8$ decagram

Hence, the mass of the released gas is 8 decagram.