First Law Of Thermodynamics MSQ - Physics MCQ

Let p_surrdV  work done by an ideal gas in the three types of expansion.
p_surr is a pressure surrounding
Since 
The maximum work is done when the gas pressure

For reversible expansion 

Path 2 corresponds to a permitted or irreversible expansion, in this case, work done

In a forbidden expansion i.e. one against an external pressure greater than the gas pressure, the work done by the gas would have to be grater than the maximum work, such an expansion is never observed.
∴ In case of isothermal expansion
Permitted process : 
Reversible process : 
Forbidden process : 
The correct answers are:

The heat has not been transferred to coffee which is thermally insulated by shaking, work has been done on coffee (system) against the viscous forces in it.
According to first law,
ΔU = Q – W
Here Q = 0
And W is negative as work is done on the system so that
ΔU = 0 – (–W)
i.e. ΔU is positive. The internal energy of system (coffee) increases.
Q = 0, W = -ve, ΔU = +ve
The correct answers are: Q = 0, W = –ve, ΔU = +ve

For an ideal gas,

Whatever be the number of atoms each gas molecule has
The correct answers are: Monatomic, Diatomic, Polyatomic

Energy changes with mass of the body, therefore, it is an extensive property.
Specific energy is the energy of the system per unit mass of the system, therefore, it will become intensive property. Heat capacity is product of specific heat and mass of the body. It depends on mass of the system,
∴ heat capacity is an extensive property. Internal energy of the system is independent of the path followed by system. It has fixed value along the path
∴ energy is the point function
The correct answers are: Energy is an extensive property, Energy is a point function, Heat capacity is an extensive property

ΔU = ΔE

Since processes is adiabatic q = 0.
As the gas expands against constant external processes

= –1.5(16–4) = –18atm-dm³
ΔU = Q + W = 0 + (–18) = –18atm-dm³
The correct answers are:  W = –18atm – dm³ , ΔU = –18atm – dm³

For one mole of an ideal gas

⇒ dU = C_VdT
For isothermal process, T is constant

Also ΔH = ΔU + Δ(pV)
pV = RT

= ΔU + RΔT
ΔH = 0.
The correct answers are: ΔU = 0, ΔH = 0

Heat transfer as well as work transfer between the system and surrounding depends upon the path by which the process is occurred. Therefore heat energy and work energy are path function. The change in internal energy ΔE remains constant, no matter which path is followed by a system to undergo a change of a certain state. Thus, internal energy is a point function or state function.
The correct answers are: Heat Energy, Work

There is no flow of heat into the resistor and work is done on the resistor (system).
According to first law as applied to the system, we write,
ΔU = Q – W
For resistor Q = 0
W = –ve. Thus
ΔU = 0 – (–W) = W
∴ Q = 0; W = –ve, ΔU = +ve
The correct answers are: Q = 0, W = –ve, ΔU = +ve

Since work is done on the system, therefore, algebraic sign of work done will be positive and magnitude of work done is 214J.
Since heat is extracted from the system, 
∴ Q = –293 J
Now ΔU = Q + W
= –293 J + 214 J
ΔU = –79 J
The correct answers are: Q = –293 J, ΔU = –79 J