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Class 11th Thermodynamics

Assertion (A): Combustion of all organic compounds is an exothermic reaction.

Reason (R) : The enthalpies of all elements in their standard state are zero.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) A is false but R is true.

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

option (ii)

Explanation: The enthalpy of the reactants is always greater than the enthalpy of the product in a combustion reaction.

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Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Four

A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill ? Take g =10 m/s².

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- speed of jackets = 125m/s

Height of hill = 500m

To cross the hill vertical component of velocity should be grater than this value u_y= $\sqrt[]{2 g h}$

$= \sqrt[]{2 * 10 * 500} = 100 m / s$

So u²= u_x²+u_y²

Horizontal component of initial velocity u_x= $\sqrt[]{u^{2} - u_{y}^{2}} = \sqrt[]{125^{2} - 100^{2}} = 75 m / s$

Time taken to reach the top of hill t= $\sqrt[]{\frac{2 h}{g}} = \sqrt[]{\frac{2 * 500}{10}} = 10 s$

Time taken to reach the ground in 10 sec = 75 (10)= 750m

Distance through which the canon has to be moved =800-750=50m

Speed with which canon can move = 2m/s

Time taken canon = 50/2= 25s

Total time t= 25+10+10= 45s

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Class 11th Thermodynamics

Match the following:

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

(i)- (b), (d)

(ii)- (b)

(iii)- (c)

(iv)- (a)

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a year ago

Class 11th Thermodynamics

Match the following parameters with description for spontaneity:

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

(i)- (c)

(ii)- (a)

(iii)- (b)

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a year ago

Class 11th Thermodynamics

Match the following processes with entropy change:

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

(i) (b)

(ii) (c)

(iii) (a)

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New answer posted

a year ago

Class 11th Thermodynamics

Match the following :

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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New answer posted

a year ago

Class 11th Thermodynamics

Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below:

2Zn(s) + O₂(g) → 2ZnO (s) ; ∆H = – 693.8 kJ/mol

(i) The enthalpy of two moles of ZnO is less than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.

(ii) The enthalpy of two moles of ZnO is more than the total enthalpy of two moles of Zn and one mole of oxygen by 693.8 kJ.

(iii) 8 kJ mol–1 energy is evolved in the reaction.

(iv) 8 kJ mol–1 energy is absorbed in the reaction.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (i) and (iii)

Since, the above reaction is an exothermic reaction, the enthalpy of reactants is more than the enthalpy of products

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Class 11th Thermodynamics

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression w= -nRTln $\frac{V_{f}}{V_{i}}$

A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively. Choose the correct option.

(i) Work done at 600 K is 20 times the work done at 300 K.

(ii) Work done at 300 K is twice the work done at 600 K.

(iii) Work done at 600 K is twice the work done at 300 K.

(iv) ∆U = 0 in both cases.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iv)

For
isothermal reversible change

q= -w = nRTln $\frac{V_{f}}{V_{i}}$ =2.303nRTlog

$\frac{W_{600 k}}{W_{300 k}}$ = $\frac{1 * R * 600 K l n \frac{10}{1}}{1 * R * 300 K l n \frac{10}{1}}$ = $\frac{600}{300}$ =2

For
isothermal expansion of ideal gases, ? U = 0

Since,
temperature is constant this means there is no change in internal energy.

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Class 11th Thermodynamics

The spontaneity means, having the potential to proceed without the assistance of external agency. The processes which occur spontaneously are (i) flow of heat from colder to warmer body. (ii) Gas in a container contracting into one corner. (iii) Gas expanding to fill the available volume. (iv) Burning carbon in oxygen to give carbon dioxide.

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iii) and (iv)

Gas expands to fill the available space spontaneously, and burning of carbon to carbon dioxide is spontaneous

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Class 11th Thermodynamics

In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. For such system

(i) qp will be negative

(ii) ∆_rH will be negative

(iii) qp will be positive

(iv) ∆_rH will be positive

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (i) and (ii)

In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and? _rH will also be negative. Hence, option (i) and (ii) is the correct answer.

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