Question

A vessel containing 1 g of oxygen at a pressure of 10 atm and a temperature of 47${}^o$C. If is found that because of a leak, the pressure drops to 5/8th of its original value and the temperature galls to 27${}^o$C. Find the volume of the vessel and the mass of oxygen that is leaked out.

• A. 0.082 L, $\frac{31}{32}$ g
• B. 0.82 L, $\frac{32}{31}$ g
• C. 8.2 L, $\frac{31}{32}$ g
• D. 0.082 L, $\frac{32}{31}$ g

Answer 1

The correct option is A 0.082 L, $\frac{31}{32}$ g

Using equation $PV=nRT$for ideal gas

$P_1=10atm=1013250Pa,n_1=\frac{1g}{/}32g/mol,T_1=273+47=320K$

$P_{1}V=n_{1}R{T}_1$...(i)

$1013250\times v=\frac{1}{32}moles\times 8.314J/mol/K\times 320K$

$\Rightarrow v=8.2\times {10}^{-}5=.082l$

After leak assuming $n_2$ moles of oxygen remained, the volume of container is constant

$P_{2}v=n_{2}R{T}_2$...(ii)

given $P=5/8P;T=273+27=300K$

dividing equation (i) by (ii)

$\frac{8}{5}=\frac{n_1{T}_1}{n_2{T}_2}$

$n_2=n_1/32=1/1024moles$

mass leaked$=(n_1-n_2)\times 32g/moles=31/32gms$