$\bullet$ Dalton's Law of partial pressure states that the total pressure by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases.
$\bullet$ ${p_{Total}} = {p_1} + {p_2} + {p_3}$
$\bullet$ Also, ${p_i} = {\chi _i}p$; where pi and $\chi$i are the partial pressure and mole fraction of ith gas respectively and p is the total pressure.
A 10.0 L flask contains 64 g of oxygen at 27$^\circ$C. (Assume O2 gas is behaving ideally). The pressure inside the flask in bar is (Given R = 0.0831 L bar K$-$1 mol$-$1)
A mixture of N2 and Ar gases in a cylinder contains 7g of N2 and 8g of Ar. If the total pressure of the mixture of the gases in the cylinder is 27 bar, the partial pressure of N2 is : [Use atomic masses (in g mol$ - $1) : N = 14, Ar = 40]
A gas at 350 K and 15 bar has molar volume 20 percent smaller than that for an ideal gas under the same
conditions. The correct option about the gas and its compressibility factor (Z) is -
A.
Z < 1 and attractive forces are dominant
B.
Z < 1 and repulsive forces are dominant
C.
Z > 1 and attractive forces are dominant
D.
Z > 1 and repulsive forces are dominant
Correct Answer: A
Explanation:
Compressibility factor (Z ) = ${{{V_{real}}} \over {{V_{ideal}}}}$
Here, Vreal < Videal. Therefore, Z < 1.
In gaseous molecules attractive forces are dominant.
Given van der Waals’ constant for NH3
, H2,
O2
and CO2
are respectively 4.17, 0.244, 1.36
and 3.59, which one of the following gases is
most easily liquefied?
A.
NH3
B.
H2
C.
O2
D.
CO2
Correct Answer: A
Explanation:
van der Waals’ constant ‘a’ signifies the
intermolecular forces of attraction between the
particle of gas. So, higher the value of ‘a’, easier
will be the liquefaction of gas.
Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for onehalf of the hydrogen to escape?
For real gases van der Waals equation is written as
$\left( {p + {{a{n^2}} \over {{V^2}}}} \right)$ (V $-$ nb) = n RT
where $a$ and $b$ are van der Waals constants. Two sets of gases are
(I) O2, CO2, H2 and He
(II) CH4. O2 and H2
The gases given in set-I in increasing order of b and gases given in set-II in decreasing order of $a$, are arranged below. Select the correct order from the following
A.
(I) He < H2 < CO2 < O2 (II) CH4 > H2 > O2
B.
(I) O2 < He < H2 < CO2 (II) H2 > O2 > CH4
C.
(I) H2 < He < O2 < CO2 (II) CH4 > O2 > H2
D.
(I) H2 < O2 < He < CO2 (II) O2 > CH4 > H2
Correct Answer: C
Explanation:
Van der Waal gas constant '$a$' represent intermolecular force of attraction of gaseous molecules and Van der Waal gas constant 'b' represent effective size of molecules . Therefore order should be
50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is 36, the molecular mass of gass A will be
A bubble of air is underwater at temperature 15oC and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25oC and the pressure is 1.0 bar, what will happen to the volume of the bubble?
In Duma's method of estimation of nitrogen 0.35 g of an organic compound gave 55 mLof nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be (aqueous tension at 300 K = 15 mm).
Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will be
A.
50.00 u
B.
12.25 u
C.
6.50 u
D.
25.00 u
Correct Answer: B
Explanation:
We know that ${{{r_A}} \over {{r_B}}} = {{v/{t_A}} \over {v/{t_B}}} = \sqrt {{{{M_B}} \over {{M_A}}}} $
A gaseous mixture was prepared by taking equal mole of CO and N2. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N2) in the mixture is
A.
0.5 atm
B.
0.8 atm
C.
0.9 atm
D.
1 atm
Correct Answer: A
Explanation:
Number of moles $ \Rightarrow $ nCO = nN2
Volume of container is same. So, VCO = VN2
.
Also,
temperature is same for both the gases thus,
TCO = TN2
According to ideal gas equation,
PV = nRT
Now, V, n, R and T for both gases are same. So,
PCO = PN2
Now, total pressure is 1 atm and according to
Dalton’s law of partial pressure,
The energy absorbed by each molecule (A2) of a substance is 4.4 $ \times $ 10$-$19 J and bond energy per molecule is 4.0 $ \times $ 10$-$19 J. The kinetic energy of the molecule per atom will be
A.
2.2 $ \times $ 10$-$19 J
B.
2.0 $ \times $ 10$-$19 J
C.
4.0 $ \times $ 10$-$20 J
D.
2.0 $ \times $ 10$-$20 J
Correct Answer: D
Explanation:
Energy absorbed by each molecule = Bond
energy per molecule + Kinetic energy per molecule
$ \Rightarrow $ 4.4 × 10–19 J = 4.0 × 10–19 J + Kinetic energy per
molecule
$ \Rightarrow $ 0.4 × 10–19 = Kinetic energy per molecule
In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture underthe aforesaid condition in the end ?
Which of the following expressions correctly
represents the relationship between the average
molar kinetic energy, KE, of CO and N2 molecules
at the same temperature ?
A.
KECO = KEN2
B.
KECO > KEN2
C.
KECO < KEN2
D.
cannot be predicted unless volumes of the
gases are given
Correct Answer: A
Explanation:
Average molar kinetic energy = ${3 \over 2}nRT$
As temperature is same hence average kinetic
energy of CO and N2 will be same.