An ideal gas is compressed to half its initial volume by means of several processes. Which of the following processes results in the maximum work being done on the gas?
1. Adiabatic
2. Isobaric
3. Isochoric
4. Isothermal

The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is -20°C, the temperature of the surroundings to which it rejects heat is  -

1. 31°C

2. 41°C

3. 11°C

4. 21°C

An ideal gas goes from state A to state B via three different processes, as indicated in the P-V diagram. If ${\mathrm{Q}}_1,$ ${\mathrm{Q}}_2,$ ${\mathrm{Q}}_3$ indicates the heat absorbed by the gas along the three processes and $∆{\mathrm{U}}_1,$ $∆{\mathrm{U}}_2,$ $∆{\mathrm{U}}_3$ indicates the change in internal energy along the three processes respectively, then:

If $∆U and ∆W$ represent the increase in internal energy and work done by the system respectively in a thermodynamical process,which of the following is true?

(1) $∆U=-∆W,$ in a adiabatic process

(2) $∆U=∆W,$ in a isothermal process

(3) $∆U=∆W,$ in adiabatic process

(4) $∆U=-∆W,$in a isothermal process 

A monoatomic gas at pressure ${\mathrm{p}}_1$ and ${\mathrm{V}}_1$ is compressed adiabatically to $\frac{1}{8}\mathrm{th}$ its original volume. What is the final pressure of the gas ?

(1) $64 {\mathrm{p}}_1$                                   

(2) ${\mathrm{p}}_1$

(3) $16 {\mathrm{p}}_1$                                   

(4) $32 {\mathrm{p}}_1$

The internal energy change in a system that has absorbed 2 kcal of heat and done 500 J of work is 

(1) 8900 J                                     

(2) 6400 J

(3) 5400 J                                     

(4) 7900 J

If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cyclic process, then

(1) W=0                                 

(2) Q=W=0

(3) E=0                                   

(4) Q=0

In a thermodynamic process, pressure of a fixed mass of a gas is changed in such a manner that the gas molecules absorb 30 J of heat and 10 J of work is done by the gas. If the initial internal energy of the gas was 40 J, then the final internal energy will be -

(1) 30 J

(2) 20 J

(3) 60 J

(4) 40 J

If the ratio of specific heat of a gas at constant pressure to that at constant volume is γ, the change in internal energy of a mass of gas, when the volume changes from V to 2V constant pressure p, is 

(1) $R/(\gamma -1)$

(2) pV

(3) $pV/(\gamma -1)$

(4) $\gamma pV/(\gamma -1)$

If heat given to a system is 6 kcal and work done by the system is 6 kJ. Then the change in internal energy is :

(1) 19.1 kJ

(2) 12.5 kJ

(3) 25 kJ

(4) Zero