Question

At a temperature, ${\text{T in K}}$, the pressure of $4{\text{ g}}$ Argon in a bulb is ${\text{P}}$. The bulb is put in a bath having temperature higher by $50{\text{ K}}$ than the first one. $0 \cdot 8{\text{ g}}$ of Argon gas had to be removed to maintain original pressure. The temperature ${\text{T}}$ is equal to:
A. $510{\text{ K}}$
B. ${\text{200 K}}$
C. $100{\text{ K}}$
D. ${\text{73 K}}$

Answer 1

Ideal gas law is an expression which relates pressure, and volume of gas with temperature of the system, and the amount of substance of the gas molecule (number of moles). It is given by the expression: ${\text{PV = nRT}}$ ,
where, c is the pressure of the gas,
-${\text{V}}$ is the volume,
-${\text{n}}$ is number of moles,
-${\text{R}}$ is the Universal gas constant,
and, ${\text{T}}$ is the temperature at Kelvin.

Complete step by step answer:
Step (1): Calculation of moles, ${\text{n}}$:
Mass of Argon gas at pressure ${\text{P}}$ is given $4{\text{ g}}$. Molar mass of an Argon is known as $40{\text{ g mo}}{{\text{l}}^{ - 1}}$.
Moles of Argon at temperature ${\text{T K}}$, ${{\text{n}}_1} = {\text{ }}\dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}$
$\Rightarrow {{\text{n}}_1}{\text{ = }}\dfrac{{4{\text{ g}}}}{{40{\text{ g}}}}{\text{ }}..........\left( {\text{1}} \right)$
It is said that $0 \cdot 8{\text{ g}}$gas is removed at temperature $\left( {{\text{T + 50}}} \right){\text{ K}}$. Then, the mass of Argon gas will be $\left( {4 - 0 \cdot 8 = 3 \cdot 2} \right){\text{ g}}$. So, moles of Argon at temperature $\left( {{\text{T + 50}}} \right){\text{ K}}$, ${{\text{n}}_2} = {\text{ }}\dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}$
$\Rightarrow {{\text{n}}_2}{\text{ = }}\dfrac{{{\text{3}} \cdot {\text{2 g}}}}{{40{\text{ g}}}}{\text{ }}..........\left( 2 \right)$

Step (2): Calculation of pressure:
According to the question, the volume of the gas is constant throughout the process, and after removing $0 \cdot 8{\text{ g}}$ gas, the final pressure becomes equal to the original pressure ${\text{P}}$. So, applying Ideal gas equation before placing the bulb in the hot bath:
${\text{PV = }}{{\text{n}}_1}{\text{RT}}$, and putting the value of ${{\text{n}}_1}$ from equation $\left( {\text{1}} \right)$,
$\Rightarrow {\text{PV = }}\dfrac{4}{{40}}{\text{RT }}.............{\text{(3)}}$
Now, applying Ideal gas equation after placing the bulb in the hot bath:
${\text{PV = }}{{\text{n}}_2}{\text{RT}}$, and putting the value of ${{\text{n}}_2}$ from equation $\left( 2 \right)$,
$\Rightarrow {\text{PV = }}\dfrac{{3 \cdot 2}}{{40}}{\text{RT }}.............{\text{(4)}}$
From equation ${\text{3 and 4}}$, $4T = 3 \cdot 2\left( {T + 50} \right)$
$\Rightarrow 4T = 3 \cdot 2T + 160$
$\Rightarrow 0 \cdot 8T = 160$
$\Rightarrow T = 200{\text{ K}}$
So, ${\text{T}}$ is equal to $200{\text{ K}}$.
Hence, option (b) is the right answer.

Note: Ideal gas equation is a combination of Boyle’s law, Charles’s law, Avogadro’s law, and Gay Lussac’s law. Among these Gay Lussac’s law can be used to interpret the above situation of the question because it gives direct relation or pressure ${\text{P}}$ with the temperature ${\text{T}}$ as $P \propto T$.