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 $3moleC{O}_2$ and $2moleofHe$ 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 $400K,$
$P={\displaystyle \frac{nRT}{V}}={\displaystyle \frac{3\times 0.0821\times 400}{8.21}}=12atm$
Pressure due to $He$ gas at $400K,$
$P={\displaystyle \frac{nRT}{V}}={\displaystyle \frac{2\times 0.0821\times 400}{8.21}}=8atm$
Pressure due to $He$ gas at $500K,$
By using Gay Lussac Law, ${\displaystyle \frac{P_1}{T_1}}={\displaystyle \frac{P_2}{T_2}}$
${\displaystyle \frac{8}{400}}={\displaystyle \frac{P_2}{500}}$
$P_2=10atm$
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 $PV=nRT$