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The number of translational degrees of freedom for a diatomic gas is [1993]- a)2
- b)3
- c)5
- d)6
Correct answer is option 'B'. Can you explain this answer?
The number of translational degrees of freedom for a diatomic gas is [1993]
a)
2
b)
3
c)
5
d)
6
|
Prisha Singh answered |
Number of translational degrees of freedom are same for all types of gases.
The degree of freedom of a molecule of a triatomic ga s is[1999]- a)2
- b)4
- c)6
- d)8
Correct answer is option 'C'. Can you explain this answer?
The degree of freedom of a molecule of a triatomic ga s is[1999]
a)
2
b)
4
c)
6
d)
8
|
Krish Khanna answered |
No. of degree of freedom = 3 K – N where K is no. of atom and N is the number of relations between atoms. For triatomic gas, K = 3, N = 3C2 = 3
No. of degree of freedom = 3 (3) – 3 = 6
No. of degree of freedom = 3 (3) – 3 = 6
[∵ PV = nRT]According to kinetic theory of gases, at absolute zero temperature [1990]- a)water freezes
- b)liquid helium freezes
- c)molecular motion stops
- d)liquid hydrogen freezes
Correct answer is option 'C'. Can you explain this answer?
According to kinetic theory of gases, at absolute zero temperature [1990]
a)
water freezes
b)
liquid helium freezes
c)
molecular motion stops
d)
liquid hydrogen freezes
|
|
Nirvikar Shakya answered |
See it is assumption only that at absolute zero temperature i.e 0 K temperature all type of molecular motion stopes since at earth it is impossible to achieve 0 K temp so i said it is mearly a assumption only . Though it is clear that all materials freeze at absolute 0 temperature so clearly water, liquid helium and liquid Hydrogen they all freezes . BUT ACCORDING TO THEORY OPTION C SEEMS TO BE MORE ACCURATE.
In a vessel, the gas is at a pressure P. If the mass of all the molecules is halved and their speed is doubled, then the resultant pressure will be [NEET Kar. 2013]- a)4P
- b)2P
- c)P
- d)P/2
Correct answer is option 'B'. Can you explain this answer?
In a vessel, the gas is at a pressure P. If the mass of all the molecules is halved and their speed is doubled, then the resultant pressure will be [NEET Kar. 2013]
a)
4P
b)
2P
c)
P
d)
P/2
|
Krish Chakraborty answered |
When mass is halved and speed is doubled then Resultant pressure,
= 2 P.
= 2 P.
The amount of heat energy required to raise the temperature of 1g of Helium at NTP, from T1K to T2K is [NEET 2013]- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The amount of heat energy required to raise the temperature of 1g of Helium at NTP, from T1K to T2K is [NEET 2013]
a)
b)
c)
d)
|
Abhishek Desai answered |
From first law of thermodynamics ΔQ = ΔU + ΔW
At 0 K, which of the following properties of a gas will be zero? [1996]- a)Kinetic energy
- b)Potential energy
- c)Density
- d)Ma ss
Correct answer is option 'A'. Can you explain this answer?
At 0 K, which of the following properties of a gas will be zero? [1996]
a)
Kinetic energy
b)
Potential energy
c)
Density
d)
Ma ss
|
Mahi Shah answered |
At 0 K, molecular motion stops. Hence, kinetic energy of molecules becomes zero.
One mole of an ideal monoatomic gas requires 207 J heat to raise the temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K, the heat required is [Given the gas constant R = 8.3 J/ mol. K] [1990]- a)198.7 J
- b)29 J
- c)215.3 J
- d)124 J
Correct answer is option 'D'. Can you explain this answer?
One mole of an ideal monoatomic gas requires 207 J heat to raise the temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same 10 K, the heat required is [Given the gas constant R = 8.3 J/ mol. K] [1990]
a)
198.7 J
b)
29 J
c)
215.3 J
d)
124 J
|
Ritika Khanna answered |
We know that Qv = nCvΔT and Qp = nCpΔT
Given Qp = 207 J 
Two containers A and B are partly filled with water and closed. The volume of A is twice that of B and it contains half the amount of water in B. If both are at the same temperature, the water vapour in the containers will have pressure in the ratio of [1988]- a)1 : 2
- b)1 : 1
- c)2 : 1
- d)4 : 1
Correct answer is option 'B'. Can you explain this answer?
Two containers A and B are partly filled with water and closed. The volume of A is twice that of B and it contains half the amount of water in B. If both are at the same temperature, the water vapour in the containers will have pressure in the ratio of [1988]
a)
1 : 2
b)
1 : 1
c)
2 : 1
d)
4 : 1
|
Lekshmi Banerjee answered |
Vapour pressure does not depend on the amount of substance. It depends on the temperature alone.
At constant volume, temperature is increased then- a)collision on walls will be less [1989]
- b)number of collisions per unit time will increase
- c)collisions will be in straight lines
- d)collisions will not change.
Correct answer is option 'B'. Can you explain this answer?
At constant volume, temperature is increased then
a)
collision on walls will be less [1989]
b)
number of collisions per unit time will increase
c)
collisions will be in straight lines
d)
collisions will not change.
|
Jatin Chakraborty answered |
As the temperature increases, the average velocity increases. So, the collisions are faster.
The pressure of a gas is raised from 27°C to 927°C.The root mean square speed is [1994]- a)
times the earlier value - b)remain the same
- c)gets halved
- d)get doubled
Correct answer is option 'D'. Can you explain this answer?
The pressure of a gas is raised from 27°C to 927°C.The root mean square speed is [1994]
a)
times the earlier valueb)
remain the same
c)
gets halved
d)
get doubled
|
Nilotpal Gupta answered |
As temperature increases from 300 K to 1200 K that is four times, so, crms will be doubled.
The equation of state, corresponding to 8g of O2 is[1994]- a)PV = 8 RT
- b)PV = RT /4
- c)PV = RT
- d)PV = RT/2
Correct answer is option 'B'. Can you explain this answer?
The equation of state, corresponding to 8g of O2 is[1994]
a)
PV = 8 RT
b)
PV = RT /4
c)
PV = RT
d)
PV = RT/2
|
Ayush Sengupta answered |
8g of oxygen is equivalent to (1/4) mole.
In the given (V – T) diagram, what is the relation between pressure P1 and P2 ? [NEET 2013]
- a)P2 > P1
- b)P2 < P1
- c)Cannot be predicted
- d)P2 = P1
Correct answer is option 'B'. Can you explain this answer?
In the given (V – T) diagram, what is the relation between pressure P1 and P2 ? [NEET 2013]
a)
P2 > P1
b)
P2 < P1
c)
Cannot be predicted
d)
P2 = P1
|
|
Nilotpal Gupta answered |
P1 > P2
As V = constant ⇒ P ∝ T
Hence from V–T graph P1 > P2
Hence from V–T graph P1 > P2
A polyatomic gas with n degrees of freedom has a mean energy per molecule given by [1989]- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A polyatomic gas with n degrees of freedom has a mean energy per molecule given by [1989]
a)
b)
c)
d)
|
Sushant Goyal answered |
According to law of equipartition of energy, the energy per degree of freedom is 
For a polyatomic gas with n degrees of freedom, the mean energy per molecule =
For a polyatomic gas with n degrees of freedom, the mean energy per molecule =
Three containers of the same volume contain three different gases. The masses of the molecules are m1, m2 and m3 and the number of molecules in their respective containers are N1, N2 and N3. The gas pressure in the containers are P1, P2 and P3 respectively. All the gases are now mixed and put in one of these containers.The pressure P of the mixture will be [1991]- a)P < (P1 + P2 + P3)
- b)
- c)P = P1 + P2 + P3
- d)P > (P1 + P2 + P3)
Correct answer is option 'C'. Can you explain this answer?
Three containers of the same volume contain three different gases. The masses of the molecules are m1, m2 and m3 and the number of molecules in their respective containers are N1, N2 and N3. The gas pressure in the containers are P1, P2 and P3 respectively. All the gases are now mixed and put in one of these containers.The pressure P of the mixture will be [1991]
a)
P < (P1 + P2 + P3)
b)
c)
P = P1 + P2 + P3
d)
P > (P1 + P2 + P3)
|
|
Krish Chakraborty answered |
According to Dalton’s law of partial pressures, we have P = P1 + P2 + P3
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