First Law Of Thermodynamic MCQ Level - 1 (Part - 1) - Physics MCQ

Internal energy of n moles of an ideal gas at temperature T is given by

where, f = degrees of freedom
= 5 for O₃ and 3 for Ar  [∴ O₂ is diatomic, Ar is monoatomic]
Hence, 
The correct answer is: 11 RT

Work done in a cyclic process = area under the graph in p-V diagram
Area = AB × BC = (2p – p) × (2V – V) = pV
∴ So, work done = pV
The correct answer is: pV

From first law of thermodynamics,
dQ = dU + dW
dQ = dU if dW = 0
Since, dQ < 0
Therefore,   or temperature will decrease
The correct answer is: the temperature will decrease

We know that heat required to raise the temperature at constant pressure is C_p.
So,  
Similarly, at constant volume C_V

or 
The correct answer is: 50

For an ideal gas : pV = nRT

For p = constant


Therefore,δ is inversely proportional to temperature T i.e., when T increases, δ decreases and vice-versa.
Hence,  graph will be rectangular hyperbola as shown in the above figure.
The correct answer is:  

The corresponding p-V graph (also called indicator diagram) in three different process will be as shown :

Area under the graph gives the work done by the gas.
(Area)₂ > (Area)₁ > (Area)₃
∴ W₂ > W₁ > W₃
The correct answer is:
W₂ > W₁ > W₃

Slope of adiabatic process at a given state (p, V, T) is more than the slope of isothermal process. The corresponding p-V graph for the two process is as shown in figure.

In the graph, AB is isothermal and BC is adiabatic.
W_AB is positive (as volume is increasing)
and W_BC is negative (as volume is decreasing)
 as area under p–V graph gives the work done.
Hence, 
From the graph itself, it is clear that p₃ > p₁
The correct answer is: p₃ > p₁ , W < 0

Heat lost by steam =Heat gained by
(water +calorimeter)