Objective Problem
Quiz by Vaijayanthi Pillai
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- Q1
First law of thermodynamics corresponds to
conservation of energy
law of conservation of angular momentum
heat flow from hotter to colder body
Newton’s law of cooling
30s - Q2
A system is taken from state A to state B along two different paths 1 and 2. The heat absorbed and work done by the system along these two paths areQ1 andQ2 and W1 and W2 , respectively. Then
Q1 +W1= Q2 +W2
W1=W2
Q1= Q2
Q 1-W1 = Q2- W2
30s - Q3
A certain amount of an ideal gas is taken from state A to state B, one time by process I and another time by process II. If the amount of heat absorbed by the gas are Q1 and Q2 respectively, then
Q1 Q2
Q1 Q 2
Q1= Q 2
None of these
30s - Q4
Work done by the gas in the process shown in figure is
positive
negative
may be positive or negative
zero
30s - Q5
Identify the incorrect statement related to a cyclic process
Q =W
W > 0
None of the above
the initial and final conditions always coincide
30s - Q6
An ideal gas of volume 1.5 X10-3 m3 ×−and at pressure1.0 105× Pa is supplied with 70 J of energy. The volume increases to 1.7 X10 - 3 m 3 ×−the pressure remaining constant. The internal energy of the gas is
increased by 90 J
increased by 70 J
increased by 50 J
decreased by 50 J
30s - Q7
Corresponding to the process shown in figure, what is the heat given to the gas in the process ABCA?
3/2 J
0
1/2 J
1 J
30s - Q8
A closed system undergoes a change of state by process 1 →2 for which Q12 = 10 J and W12 = − 5 J. The system is now returned to its initial state by a different path 2 →1 for which Q21 is −3 J. The work done by the gas in the process 2 → 1 is
−2 J
+5 J
zero
8 J
30s - Q9
A thermodynamical system is changed from state (p1, V 1) to ( p2, V 2 ) by two different process. The quantity which will remain same will be?
∆W
∆Q- ∆W
∆Q+ ∆ W
∆Q
30s - Q10
In thermodynamic process, 200 J of heat is given to a gasand 100 J of work is also done on it. The change in internal energy of the gas is :
100 J
419 J
300 J
240 J
30s - Q11
If ∆Q and ∆W 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
∆Q= ∆U - ∆ W
∆Q =∆W- ∆U
∆Q= ∆U + ∆ W
∆Q =-∆W- ∆U
30s - Q12
In a given process for an ideal gas, dW = 0 and dQ < 0.Then, for the gas
the temperature will decrease
the temperature will increase
the volume will increase
the pressure will remain constant
30s - Q13
When 1 g of water at 0°C and 1 X105 N / m 2 × pressure is converted into ice of volume 1.092 cm3, then work done will be
Data insufficient
0.091 J
0.0182 J
0.0091 J
30s - Q14
In an adiabatic expansion of one mole of gas initial and final temperatures are T1and T2 respectively, then the change in internal energy of the gas is
zero
R (T1- T2)
R/Y-1 (T2- T1)
R/Y-1 (T1- T2)
30s - Q15
A gas expands under constant pressure p from volumeV1andV2. The work done by the gas is
p (V1- V2)
p (V1 - V2)
p (V2- V1)
p V1 V2/V2- V1
30s
