JEE Advanced Previous Year Questions (2021 - 2024): Gaseous State | Chemistry for JEE Main & Advanced PDF Download

 Page 1


 
Q1: A closed vessel contains 10 g of an ideal gas X at 300 K, which exerts 2 atm pressure. At the
same temperature, 80 g of another ideal gas Y is added to it and the pressure becomes 6 atm.
The ratio of root mean square velocities of X and Y at 300 K is
(a) 2 v2 : v3
(b) 2 v2 : 1
(c) 1 : 2
(d) 2 : 1    [JEE Advanced 2024 Paper 1]
Ans: (d)
T o solve this problem, we'll start by using the ideal gas law and the formula for the root mean square
(rms) velocity of a gas.
The ideal gas law is given by:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant,
and T is the temperature. The rms velocity (v ) of a gas is given by:
v = v(3RT/M)
where M is the molar mass of the gas. We need the rms velocities of gases X and Y.
First, let's nd the number of moles of each gas. For gas X, let’s denote the molar mass as M . The
number of moles is:
n = 
10 g
M
For the initial gas X at 300 K and 2 atm, we can write:
P V = n RT
When gas Y is added, the total pressure becomes 6 atm. Let the molar mass of Y be M . The number
of moles of gas Y added is:
n = 
80 g
M
rms
rms
X
X
X
X X
Y
Y
Y
Page 2


 
Q1: A closed vessel contains 10 g of an ideal gas X at 300 K, which exerts 2 atm pressure. At the
same temperature, 80 g of another ideal gas Y is added to it and the pressure becomes 6 atm.
The ratio of root mean square velocities of X and Y at 300 K is
(a) 2 v2 : v3
(b) 2 v2 : 1
(c) 1 : 2
(d) 2 : 1    [JEE Advanced 2024 Paper 1]
Ans: (d)
T o solve this problem, we'll start by using the ideal gas law and the formula for the root mean square
(rms) velocity of a gas.
The ideal gas law is given by:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant,
and T is the temperature. The rms velocity (v ) of a gas is given by:
v = v(3RT/M)
where M is the molar mass of the gas. We need the rms velocities of gases X and Y.
First, let's nd the number of moles of each gas. For gas X, let’s denote the molar mass as M . The
number of moles is:
n = 
10 g
M
For the initial gas X at 300 K and 2 atm, we can write:
P V = n RT
When gas Y is added, the total pressure becomes 6 atm. Let the molar mass of Y be M . The number
of moles of gas Y added is:
n = 
80 g
M
rms
rms
X
X
X
X X
Y
Y
Y
Now, the total number of moles in the vessel is n + n , and this total exerts a pressure of 6 atm at
the same volume and temperature:
P V = (n + n )RT
Solving for the volume V and equating, we get:
2V = 
10
M
 RT
6V = 
10
M
 + 
80
M
 RT
Dividing and solving for 
80
M
, we get:
80
M
 = 2 
10
M
M = 4M
Now, applying the formula for the rms velocity, the ratio of rms velocities of X and Y is:
v
v
 = v 
M
M
 = v 
4M
M
 = 2
So, the ratio of root mean square velocities of X and Y is 2 : 1.
Thus, the correct answer is Option D: 2 : 1.
X Y
total X Y
X
X Y
Y
Y X
Y X
rms,X
rms,Y
Y
X
X
X
Page 3


 
Q1: A closed vessel contains 10 g of an ideal gas X at 300 K, which exerts 2 atm pressure. At the
same temperature, 80 g of another ideal gas Y is added to it and the pressure becomes 6 atm.
The ratio of root mean square velocities of X and Y at 300 K is
(a) 2 v2 : v3
(b) 2 v2 : 1
(c) 1 : 2
(d) 2 : 1    [JEE Advanced 2024 Paper 1]
Ans: (d)
T o solve this problem, we'll start by using the ideal gas law and the formula for the root mean square
(rms) velocity of a gas.
The ideal gas law is given by:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant,
and T is the temperature. The rms velocity (v ) of a gas is given by:
v = v(3RT/M)
where M is the molar mass of the gas. We need the rms velocities of gases X and Y.
First, let's nd the number of moles of each gas. For gas X, let’s denote the molar mass as M . The
number of moles is:
n = 
10 g
M
For the initial gas X at 300 K and 2 atm, we can write:
P V = n RT
When gas Y is added, the total pressure becomes 6 atm. Let the molar mass of Y be M . The number
of moles of gas Y added is:
n = 
80 g
M
rms
rms
X
X
X
X X
Y
Y
Y
Now, the total number of moles in the vessel is n + n , and this total exerts a pressure of 6 atm at
the same volume and temperature:
P V = (n + n )RT
Solving for the volume V and equating, we get:
2V = 
10
M
 RT
6V = 
10
M
 + 
80
M
 RT
Dividing and solving for 
80
M
, we get:
80
M
 = 2 
10
M
M = 4M
Now, applying the formula for the rms velocity, the ratio of rms velocities of X and Y is:
v
v
 = v 
M
M
 = v 
4M
M
 = 2
So, the ratio of root mean square velocities of X and Y is 2 : 1.
Thus, the correct answer is Option D: 2 : 1.
X Y
total X Y
X
X Y
Y
Y X
Y X
rms,X
rms,Y
Y
X
X
X
 
 
Numerical Type 
JEE Advanced 2023 Paper 1 Online 
Q1. A gas has a compressibility factor of 0.5 and a molar volume of ?? . ?? ????
?? ?? ???? - ?? at a 
temperature of ?? ?? ?? ?? and pressure ?? atm. If it shows ideal gas behaviour at the same 
temperature and pressure, the molar volume will be ????
?? ?? ?? ?? - ?? . The value of ?? / ?? is 
[Use: Gas constant, ?? = ?? × ????
- ?? ?? ?? ?? ?? ?? - ?? ?? ?? ?? - ?? ] 
Ans: 100 
For gas : Z = 0 . 5 , V
m
= 0 . 4 L / m o l 
T = 800 K , P = X atm.  
? Z =
PV
m
RT
 
?
X ( 0 . 4 )
0 . 08 × 800
= 0 . 5 
? X = 8 0 
For ideal gas, PV
m
= RT 
? V
m
=
RT
P
=
0 . 08 × 800
80
= 0 . 8 L m o l
- 1
= y 
Then, 
?? ?? =
80
0 . 8
= 100.