Ask & Answer: India's Largest Education Community

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

The presence of reactants in a closed vessel made of conducting material e.g., copper or steel is an example of a closed system.

Kindly go through the solution

This is a Multiple Choice Questions as classified in NCERT Exemplar

Thermodynamics is not concerned about how and at what rate these energy transformations are carried out but is based on initial and final states of a system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state.

$I\; ={\displaystyle \int \frac{1}{cos(x+a)cos(x+b)}dx}$

$=\frac{1}{sin(a-b)}\int \frac{sin(a-b)}{cos(x+a)cos(x+b)}·dx$

$=\frac{1}{sin(a-b)}\int \frac{sin[(x+a)-(x+b)]}{cos(x+a)cos(x+b)}dx$

$=\frac{1}{sin(a-b)}·\int \frac{sin(x+a)cos(x+b)-cos(x+a)sin(x+b)}{cos(x+a)cos(x+b)}dx$

$=\frac{1}{sin(a-b)}\left[{\displaystyle \int tan(x+a)dx}-{\displaystyle \int tan}(x+b)dx\right]$

$=\frac{1}{sin(a-b)}[-log|cos(x+a)|+log|cos(x+b)|]+\; C$

$=\frac{1}{sin(a-b)}log\left|\frac{cos(x+b)}{cos(x+a)}\right|+\; C$

$\int \frac{si{n}^{8}x-co{s}^{8}x}{1-2si{n}^{2}xco{s}^{2}x}dx=\int \frac{\left(si{n}^{4}x+co{s}^{4}x\right)\left(si{n}^{4}x-co{s}^{4}x\right)}{\left(si{n}^{2}x+co{s}^{2}x\right)-si{n}^{2}xco{s}^{2}x-si{n}^{2}xco{s}^{2}x}dx$

$=\int \frac{\left(si{n}^{4}x+co{s}^{4}x\right)\left(si{n}^{2}x+co{s}^{2}x\right)\left(si{n}^{2}x-co{s}^{2}x\right)}{si{n}^{2}x\left(1-co{s}^{2}x\right)+co{s}^{2}x\left(1-si{n}^{2}x\right)}dx\{\begin{array}{cc}\hfill ?a^2-b^2=(a+b)(a-b)\hfill & \hfill 1=si{n}^{2}x+co{s}^{2}x.\hfill \end{array}$

$={{\displaystyle \int }}^{\text{?}}\frac{\left(si{n}^{4}x+co{s}^{4}x\right)\left(si{n}^{2}x+co{s}^{4}x\right)(si{n}^{2}x-co{s}^{2}x)\times 1}{si{n}^{2}x\cdot si{n}^{2}x+co{s}^{2}x\cdot co{s}^{2}x}dx$

$=\int \frac{\left(si{n}^{4}x+co{s}^{2}x\right)\left(si{n}^{2}x-co{s}^{2}x\right)}{\left(sin4x+co{s}^{4}x\right)}dx$

$={\displaystyle \int -}\left(co{s}^{2}x-si{n}^{2}x\right)dx$

$=-{\displaystyle \int cos}2xdx$

$=-\frac{sin2x}{2}+\; C$

This is a Short Answer Type Questions as classified in NCERT Exemplar

We know that the amount of work done =-p_ext? V

On substituting the values in the formula, we get,

-2bar* (50-10)L=-80Lbar

According to the described problem,1 LBar = 100J

Therefore, -80 L bar= (-80*100)= -8000J

= -8kJ, which is the amount of work done

The significance of the negative sign states that the work is done on the surroundings of the system. In the case of reversible expansion, the work done will be more.

Kindly go through the solution

For admission to the BVoc course at PW Institute of Innovation, Noida, students need to fulfil the minimum eligibility criteria in the first place. The selection criteria for this course are entrance-based and candidates must pass Class 12. Candidates who have not completed Class 12 are not eligible for admission

This is a Short Answer Type Questions as classified in NCERT Exemplar

We can conclude from the figure that this change is a reversible change.

Now,

W= -2.303nRT log $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

But,  p₁V₁ = p₂V_{2 = $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$} = $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$ = $\frac{2}{1}$ =2

 W= -2.303nRT log $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$

 = -2.303 *8.314*1*298*log2

= -2.303 *8.314*298*0.3010J

= -1717.46J

This is a Short Answer Type Questions as classified in NCERT Exemplar

No, for the state of spontaneity, the enthalpy change is not the only criteria. Entropy also needs to be taken into account here.

$\int \frac{e^{5logx}-e^{4logx}}{e^{3logx}-e^{2logx}}dx=\int \frac{e^{log{x}^5}-e^{log{x}^4}}{e^{log{x}^3}-e^{log{x}^2}}dx\{\begin{array}{cc}\hfill ?nlogm\hfill & \hfill =log{m}^n\hfill \end{array}$

$=\int \frac{x^5-x^4}{x^3-x^2}dx\{?e^{logx}=x$

$=\int \frac{x^4(x-1)}{x^2(x-1)}dx=\int x^{2}dx=\frac{x^3}{3}+\; C$

Yes, applicants with accommodation are eligible to extend their SAT exam duration only if they have prior permission. The duration extension will depend on the applicant's need.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Throwing a stone from ground to roof

b) the reaction involved is a process where the energy decreases after the reaction. It can be represented as:In process b), potential energy/enthalpy change is a contributing factor to the spontaneity.

$=\int \frac{sinx}{sin(x-a)}dx$

Let x- a = t=> dx = dt.

$I\; =\int \frac{sin(t+a)}{sint}dt=\int \frac{sintcosa+costsina}{sint}dt$

$=cosa{\displaystyle \int d}t+sina{\displaystyle \int cot}tdt$

$=tcosa+sina·log\left|sint\right|+\; C$

$=(x-a)cosa+sinalog|sin(x-a)|+\; C$

$=xcosa+sinalog|sin(x-a)|+{}_{\mathrm{C1}}$

Where C₁ = C - a cos a

This is a Short Answer Type Questions as classified in NCERT Exemplar

When a process can be reversed by bringing an extremely small change in it, we call it a reversible process. The pressure-volume graph can be used to calculate the work done. The pressure is not constant, and changes in infinitesimal amounts as compression happens from initial volume V_i to the final volume V_f. The below graph depicts the work done with the shaded area.

From (1), B = - A = $\left(-\frac{1}{2}\right)=\frac{1}{2}$

From (2) C = 5 - B = 5 $-\frac{1}{2}=\frac{10-1}{2}=\frac{9}{2}$

$\int \frac{5x}{(x+1)\left(x^2+9\right)}dx=\int \left\{\frac{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{x+1}+\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.x+\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{x^2+9}\right\}dx$

$=-\frac{1}{2}\int \frac{dx}{x+1}+\frac{1}{2}\int \frac{x}{x^2+9}dx+\frac{9}{2}\int \frac{dx}{x^2+9}$

$=-\frac{1}{2}log|x+1|+\frac{1}{4}\int \frac{9x}{x^2+9}dx+\frac{1}{2}-\int \frac{dx}{x^2+3^2}.$

$=-\frac{1}{2}log|x+1|+\frac{1}{4}log\left(x^2+9\right)+\frac{9}{2}\times \frac{1}{3}ta{n}^{-1}\frac{x}{3}+\; C$

$=-\frac{1}{2}log|x+1|+\frac{1}{4}log\left(x^2+9\right)+\frac{3}{2}ta{n}^{-1}\frac{x}{3}+\; C$