A thermo-dynamical system is changed from state (P₁, V₁) to (P₂, V₂) by two different process. The quantity which will remain same will be

(1) ΔQ

(2) ΔW

(3) ΔQ + ΔW

(4) ΔQ – ΔW

If ΔQ and ΔW represent the heat supplied to the system and the work done on the system respectively, then the first law of thermodynamics can be written as 

(1) ΔQ = ΔU + ΔW

(2) ΔQ = ΔU – ΔW

(3) ΔQ = ΔW – ΔU

(4) ΔQ = –ΔU – ΔW

where ΔU is the internal energy

Which of the following can not determine the state of a thermodynamic system?

Which of the following is not a thermodynamics co-ordinate ?

(1) P

(2) T

(3) V

(4) R

The second law of thermodynamics states that in a cyclic process:

1. Work cannot be converted into heat

2. Heat cannot be converted into work

3. Work cannot be completely converted into heat

4. Heat cannot be completely converted into work

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

First law of thermodynamics is a special case of 

(1) Newton's law

(2) Law of conservation of energy

(3) Charle's law

(4) Law of heat exchange

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

A monoatomic gas of n-moles is heated from temperature T₁ to T₂ under two different conditions (i) at constant volume and (ii) at constant pressure. The change in internal energy of the gas is 

(1) More for (i)

(2) More for (ii)

(3) Same in both cases

(4) Independent of number of moles

Can two isothermal curves cut each other

(1) Never

(2) Yes

(3) They will cut when temperature is 0°C

(4) Yes, when the pressure is critical pressure