NEET Exam  >  NEET Tests  >  Test: Basic Concepts of Thermodynamics & Applications (June 12) - NEET MCQ

Test: Basic Concepts of Thermodynamics & Applications (June 12) - NEET MCQ


Test Description

10 Questions MCQ Test - Test: Basic Concepts of Thermodynamics & Applications (June 12)

Test: Basic Concepts of Thermodynamics & Applications (June 12) for NEET 2024 is part of NEET preparation. The Test: Basic Concepts of Thermodynamics & Applications (June 12) questions and answers have been prepared according to the NEET exam syllabus.The Test: Basic Concepts of Thermodynamics & Applications (June 12) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Basic Concepts of Thermodynamics & Applications (June 12) below.
Solutions of Test: Basic Concepts of Thermodynamics & Applications (June 12) questions in English are available as part of our course for NEET & Test: Basic Concepts of Thermodynamics & Applications (June 12) solutions in Hindi for NEET course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Basic Concepts of Thermodynamics & Applications (June 12) | 10 questions in 20 minutes | Mock test for NEET preparation | Free important questions MCQ to study for NEET Exam | Download free PDF with solutions
Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 1

Direction (Q. Nos. 1-15) This section contains 15 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q. 
Assign the sign of work done (based on SI convention) in the following chemical changes taking place against external atmospheric pressure :


    

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 1

Work done = - (change in moles) RT For work done positive change in moles negative and vice versa We only cosider the moles that are in gass phase.

Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 2

Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of XY2 weighs 10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and Y are

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 2

Let atomic weight of x = Mx
atomic weight of y = M

we know, 
mole = weight /atomic weight 
a/c to question, 

mole of xy2 = 0.1 
so, 
0.1 = 10g/( Mx +2My

Mx + 2My = 100g -------(1)

for x3y2 ; mole of x3y2 = 0.05 

0.05 = 9/( 3Mx + 2My

3Mx + 2My = 9/0.05 = 9 × 20 = 180 g ---(2) 
solve eqns (1) and (2)

2Mx = 80 
Mx = 40g/mol 

and My = 30g/mole

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 3

56 g of iron reacts with dilute H2SO4 at 27° C . W ork done (in cals) in

I. closed vessel of fixed volume and
II. an open vessel is

    

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 3

Use ∆H=∆U+∆NgRT FOR 1. we need to find work done in constant volume that is ∆U which will be 0 as it is nCv∆T and since ∆T is zero so ∆U is zero .
For 2. we need to Find ∆H =∆U+∆NgRT Since ∆U is zero so∆H=∆NgRT 
R=2 T =300and ∆Ng is = -1

Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 4

Consider the following properties. 

State functions are 

[IITJEE2009]

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 4

First 4 are fundamental examples of state functions as they are path independent. Since reversible expansion depends on the path followed by process, so it is a path function. However, irreversible expansion work is independent of the path and so, it is a state function.

Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 5

One mole of an ideal gas is put through a series of changes as shown in the figure in which 1,2,3 mark the three stages of the system. Pressure at the stages 1, 2, and 3 respectively will be (in bar)

   

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 5

Applying ideal gas eqn. at 1
p×22.4 = 1×0.0821×298        (since vol. Is in L. R should haave constaant with lit in it.)
p = 1.09 atm. To convert it into bar, we multiply it by 1.01 or we have 1.03 bar
Similarly at 2 and 3, we get value of pressure at 2 and 3. 
The most closest answer is option d

Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 6

One mole each of CaC2, AI4C3 and Mg2C3 reacts with H2O in separate open flasks at 25° C. Numerical value of the work done by the system is in order

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 6

i) CaC2 + 2H2O  →  C2H2(g) + Ca(OH)2
ii) Mg2C3 + 4H2O → CH3-C≡CH(g) + 2Mg(OH)2
iii) Al4C3 + 12H2O → 3CH4(g) + 4Al(OH)3
Work done = -∆ngRT 
Or W ∝ ∆ng
Wecan see that in case iii) we have maximum no of moles in the product side(∆ng = 3). So work done will be maximum in case iii). After that in i) and ii), we have the same number of moles on the product side(∆ng = 1), so work done will be the same.
Therefore, option c is correct.

Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 7

The internal energy change when a system goes from state A to B is 40kJ/mole. If the system goes from A to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy?

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 7


We know that for a cyclic process the net change in internal energy is equal to zero and change in the internal energy does not depend on the path by which the final state is reached.

*Answer can only contain numeric values
Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 8

A sample of gas changes from p1 V1 and T1, to p2, V2 and T2 by one path and then back to  and T1, by another path. How many of the following must be zero for the gas in this cycle?


Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 8

∆T, ∆p, ∆V and ∆U = 0 (also they are state function)
∆U = q+w. So, q and w are not zero.
Hence we have 4 parameters equal to zero after the whole process.

*Answer can only contain numeric values
Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 9

What will be the temperature change of 1.00 mole of CH3OH(g) if 100 J of heat is added to it under constant volume conditions?


Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 9

For heat added at constant volume, Work doe = 0
∆U = T1T2 nCv,mdT
Or ∆U = nCv,m∆T
∆T = ∆U/nCv,m = 100/1×33.33 = 3
So, the correct answer is 3

Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 10

The following diagram represents the (p-V) changes of a gas.

Total work done is

Detailed Solution for Test: Basic Concepts of Thermodynamics & Applications (June 12) - Question 10

Work done is the area enclosed by PV graph with volume axis.

Information about Test: Basic Concepts of Thermodynamics & Applications (June 12) Page
In this test you can find the Exam questions for Test: Basic Concepts of Thermodynamics & Applications (June 12) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Basic Concepts of Thermodynamics & Applications (June 12), EduRev gives you an ample number of Online tests for practice

Top Courses for NEET

Download as PDF

Top Courses for NEET