A balloon contains $500 {\mathrm{m}}^3$ of helium at 27°C and 1 atmosphere pressure. The volume of the helium at – 3°C temperature and 0.5 atmosphere pressure will be :

1.  $500 {\mathrm{m}}^3$                                 

2.  $700 {\mathrm{m}}^3$

3.  $900 {\mathrm{m}}^3$                                 

4. $1000 {\mathrm{m}}^3$

A vessel contains 1 mole of ${\mathrm{O}}_2$ gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at temperature 2T has a pressure of

1. P/8                                   

2. P

3. 2P                                   

4. 8P

A given mass of a gas is allowed to expand freely until its volume becomes double. If ${\mathrm{C}}_{\mathrm{b}}$ and ${\mathrm{C}}_{\mathrm{a}}$ are the velocities of sound in this gas before and after expansion respectively, then ${\mathrm{C}}_{\mathrm{a}}$ is equal to :

1. $2{\mathrm{C}}_{\mathrm{b}}$                                 

2. $\sqrt{2}{\mathrm{C}}_{\mathrm{b}}$

3. ${\mathrm{C}}_{\mathrm{b}}$                                   

4. $\frac{1}{\sqrt{2}}{\mathrm{C}}_{\mathrm{b}}$

One litre of Helium gas at a pressure 76 cm of Hg and temperature 27°C is heated till its pressure and volume are doubled. The final temperature attained by the gas is :

1.  927°C                                 

2.  900°C

3.  627°C                                 

4.  327°C

The molecular weight of a gas is 44. The volume occupied by 2.2 g of this gas at $0°\mathrm{C}$ and 2 atm pressure will be-

1.  0.56 litre                           

2.  1.2 litres

3.  2.4 litres                           

4.  5.6 litres

A gas at \(27^{\circ}\text{C}\) temperature and \(30\) atmospheric pressure is allowed to expand to the atmospheric pressure. If the volume becomes \(10\) times its initial volume, then the final temperature becomes:
1. \(100^{\circ}\text{C}\)
2. \(173^{\circ}\text{C}\)
3. \(273^{\circ}\text{C}\)
4. \(-173^{\circ}\text{C}\)

In the relation $\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}, \mathrm{n}=$

1.  Number of molecules                             

2.  Atomic number

3.  Mass number                                         

4.  Number of moles

The equation of state corresponding to 8 g of ${\mathrm{O}}_2$ is :

1.  PV = 8RT                           

2.  PV = RT/4

3.  PV = RT                             

4.  PV = RT/2

The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be

1. $\mathrm{PV}=\left(5/32\right)\mathrm{RT}$                               

2. $\mathrm{PV}=5\mathrm{RT}$

3. $\mathrm{PV}=\left(5/2\right)\mathrm{RT}$                                 

4. $\mathrm{PV}=\left(5/16\right)\mathrm{RT}$

(where R is the gas constant)

Three containers of the same volume contain three different gases. The masses of the molecules are ${\mathrm{m}}_1, {\mathrm{m}}_2$ and ${\mathrm{m}}_3$ and the number of molecules in their respective containers are ${\mathrm{N}}_1, {\mathrm{N}}_2$ and ${\mathrm{N}}_3$ . The gas pressure in the containers are ${\mathrm{P}}_1, {\mathrm{P}}_2$ and ${\mathrm{P}}_3$ respectively. All the gases are now mixed and put in one of the containers. The pressure P of the mixture will be :

1. $\mathrm{P}<\left({\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3\right)$                               

2. $\mathrm{P}=\frac{{\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3}{3}$

3. $\mathrm{P}={\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3$                                 

4. $\mathrm{P}>\left({\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3\right)$