Question

$4\text{g}$ of argon (Atomic mass = 40) in a bulb at a temperature of $TK$ has a pressure $P$ atm. When the bulb was placed in a hot bath at a temperature ${50}^{\circ }C$ more that the first one, $0.8\text{g}$ of the gas had to be removed to get the original pressure. $T$ is equal to:

• A. $510\text{K}$
• B. $200\text{K}$
• C. $100\text{K}$
• D. $73\text{K}$

Answer 1

The correct option is B $200\text{K}$

4 g of argon gas is to be maintained at the same temperature and pressure i.e., when temperature is raised, excess argon gas is removed.
So using Ideal gas equation,
PV = nRT
PV is conatant in both the conditions
So, $n_{1}R{T}_1=n_{2}R{T}_2$
$n_1=\frac{4}{40}$ = 0.1 mole
$n_2=\frac{3.2}{40}$ = 0.08 moles
initial temperature is $T_1$ K and Final temperature is $T_1+50$ K
equating all values
$0.1\times T_1=0.08\times (T_1+50)$

solving we get $T_1$ = 200 K