Question

Two closed vessels A and B of equal volume of $8.21L$ are connected by a narrow tube of negligible volume with an open valve. The left hand side container is found to contain $3\,mole\,C{O_2}$ and $2\,mole\,of\,He$ at $400K$. What is the partial pressure of $He$ in vessel B at $500K?$
A. 2.4atm
B. 8 atm
C. 12 atm
D. None of these

Answer 1

Hint: In this problem use the Ideal gas law which states that for an ideal gas $PV = nRT$ where
$P = $ Pressure,
$V = $ Volume in litres
$n = $ no. of moles
$R = $ Gas Constant (0.0821)
$T = $ Temp. in Kelvin

Complete step-by-step answer:
Pressure due to $C{O_2}$ gas at $400K,$
$P = \dfrac{{nRT}}{V} = \dfrac{{3 \times 0.0821 \times 400}}{{8.21}} = 12atm$
Pressure due to $He$ gas at $400K,$
$P = \dfrac{{nRT}}{V} = \dfrac{{2 \times 0.0821 \times 400}}{{8.21}} = 8atm$
Pressure due to $He$ gas at $500K,$
By using Gay Lussac Law, $\dfrac{{{P_1}}}{{{T_1}}} = \dfrac{{{P_2}}}{{{T_2}}}$
$\dfrac{8}{{400}} = \dfrac{{{P_2}}}{{500}}$
${P_2} = 10atm$
Thus option D “None of these” is the correct answer to the problem.

Note: According to the ideal gas equation- At low pressure and high temperature, P is inversely proportional to V (Boyle’s law), V is directly proportional to n and P is directly proportional to T (Boyle’s law).
So on solving it we get $PV = nRT$