AP EAMCET 2018 Chemistry Question Paper with Answer and Solution

(D) For an isothermal reversible expansion,the work done by the gas is given by the formula: $W = -nRT \ln(\frac{V_2}{V_1})$.
Since the process is isothermal,$p_1 V_1 = nRT$.
Given $p_1 = 20 \ atm$ and $V_1 = 1.5 \ L$,we have $nRT = 20 \ atm \times 1.5 \ L = 30 \ L \ atm$.
Substituting the values into the work formula:
$W = -30 \times \ln(\frac{15}{1.5})$
$W = -30 \times \ln(10)$
$W = -30 \times 2.303 \times \log_{10}(10)$
$W = -30 \times 2.303 \times 1 = -69.09 \ L \ atm$.
The negative sign indicates that work is done by the system on the surroundings.

(B) For an isothermal reversible expansion of an ideal gas,the work done $(w)$ is given by the formula: $w = -2.303 \ nRT \ \log(\frac{V_f}{V_i})$.
Since $PV = nRT$,we can substitute $nRT$ with $P_i V_i$ (initial state):
$w = -2.303 \ P_i V_i \ \log(\frac{V_f}{V_i})$.
Given: $w = -184.24 \ L \ atm$,$P_i = 10 \ atm$,$V_i = 4 \ L$.
Substituting the values:
$-184.24 = -2.303 \times (10 \times 4) \times \log(\frac{V_f}{4})$.
$-184.24 = -2.303 \times 40 \times \log(\frac{V_f}{4})$.
$-184.24 = -92.12 \times \log(\frac{V_f}{4})$.
$\log(\frac{V_f}{4}) = \frac{-184.24}{-92.12} = 2$.
$\frac{V_f}{4} = 10^2 = 100$.
$V_f = 100 \times 4 = 400 \ L$.

(A) The formation reaction for $CH_3OH_{(l)}$ is:
$C_{(graphite)} + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow CH_3OH_{(l)}$
We can obtain this by combining the given equations:
$(iii) + 2 \times (ii) - (i)$
Substituting the values:
$\Delta_fH^{\circ} = (-393) + 2 \times (-286) - (-726)$
$\Delta_fH^{\circ} = -393 - 572 + 726$
$\Delta_fH^{\circ} = -965 + 726 = -239 \ kJ \ mol^{-1}$

(A) The combustion reaction is: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$,$\Delta H = -394.0 \ kJ \ mol^{-1}$.
The molar mass of $CO_2$ is $12 + (2 \times 16) = 44 \ g \ mol^{-1}$.
Since the formation of $44 \ g$ of $CO_2$ releases $394.0 \ kJ$ of heat,the enthalpy change for the formation of $17.6 \ g$ of $CO_2$ is calculated as:
$\Delta H = \frac{-394.0 \ kJ \ mol^{-1} \times 17.6 \ g}{44 \ g \ mol^{-1}} = -157.6 \ kJ$.

(B) The relationship between Gibbs energy change,enthalpy change,and entropy change is given by: $\Delta G = \Delta H - T\Delta S$.
We know that $\Delta H = \Delta U + \Delta n_g RT$.
For the reaction $X_2O_{4(l)} \rightarrow 2XO_{2(g)}$,the change in the number of moles of gas is $\Delta n_g = 2 - 0 = 2$.
Given $\Delta U = x \ kJ \ mol^{-1} = 1000x \ J \ mol^{-1}$ and $\Delta S = y \ J \ K^{-1} \ mol^{-1}$.
Substituting these values into the enthalpy equation: $\Delta H = 1000x + 2RT$.
Now,substituting $\Delta H$ into the Gibbs energy equation: $\Delta G = (1000x + 2RT) - T(y)$.
Rearranging the terms,we get: $\Delta G = 1000x + T(2R - y) \ J \ mol^{-1}$.

(B) The relationship between Gibbs energy change,enthalpy change,and entropy change is given by $\Delta G = \Delta H - T\Delta S$.
Enthalpy change is related to internal energy change by $\Delta H = \Delta U + \Delta n_gRT$.
For the reaction $X_2O_{4(l)} \rightarrow 2XO_{2(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = 2 - 0 = 2$.
Substituting $\Delta H$ into the Gibbs equation: $\Delta G = \Delta U + \Delta n_gRT - T\Delta S$.
Given $\Delta U = x \ kJ \ mol^{-1} = 1000x \ J \ mol^{-1}$ and $\Delta S = y \ J \ K^{-1} \ mol^{-1}$,we get:
$\Delta G = 1000x + 2RT - Ty$.
Factoring out $T$,we obtain: $\Delta G = 1000x + T(2R - y) \ J \ mol^{-1}$.

(D) The dimensional formula for energy per unit volume is $[M L^{-1} T^{-2}]$.
The dimensional formula for angular momentum is $[M L^2 T^{-1}]$.
Since these two quantities have different dimensions,they cannot be added or subtracted. Thus,Assertion $(A)$ is false.
According to the principle of homogeneity of dimensions,only physical quantities having the same dimensions can be added or subtracted. Thus,Reason $(R)$ is true.
Therefore,the correct option is $(d)$.

(D) The linear velocity of the source,$v_s = r \omega$.
Given $r = \frac{100}{\pi} ~cm = \frac{1}{\pi} ~m$ and angular frequency $\omega = 2 \pi f_{rot} = 2 \pi \times 5 = 10 \pi ~rad/s$.
Thus,$v_s = \frac{1}{\pi} ~m \times 10 \pi ~rad/s = 10 ~m/s$.
Using the Doppler effect formula for a moving source and stationary observer:
$f' = f \left( \frac{v}{v \mp v_s} \right)$.
The maximum frequency occurs when the source moves towards the listener:
$f_{max} = 500 \times \left( \frac{332}{332 - 10} \right) = 500 \times \frac{332}{322} \approx 515.5 ~Hz$.
The minimum frequency occurs when the source moves away from the listener:
$f_{min} = 500 \times \left( \frac{332}{332 + 10} \right) = 500 \times \frac{332}{342} \approx 485.4 ~Hz$.

(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
From this,we can see that $n \propto \frac{\sqrt{T}}{l}$.
Let the initial length be $l$ and initial tension be $T$. The final length is $l' = l - \frac{x}{100}l = l(1 - \frac{x}{100})$ and the final tension is $T' = T + 0.44T = 1.44T$.
The ratio of the final frequency $n'$ to the initial frequency $n$ is given as $\frac{n'}{n} = \frac{2}{1}$.
Using the proportionality relation: $\frac{n'}{n} = \sqrt{\frac{T'}{T}} \times \frac{l}{l'}$.
Substituting the values: $\frac{2}{1} = \sqrt{\frac{1.44T}{T}} \times \frac{l}{l(1 - \frac{x}{100})}$.
$2 = \sqrt{1.44} \times \frac{1}{1 - \frac{x}{100}}$.
$2 = 1.2 \times \frac{100}{100 - x}$.
$2(100 - x) = 120$.
$200 - 2x = 120$.
$2x = 80$.
$x = 40$.

(D) Let the mass of the shell be $M$. At the highest point,the velocity of the shell is $v_x = u \cos \theta$. The shell breaks into two equal parts of mass $m = M/2$.
Let the velocity of the first part (which retraces its path) be $v_1 = -u \cos \theta$. By the law of conservation of linear momentum:
$M(u \cos \theta) = m v_1 + m v_2$
$M(u \cos \theta) = (M/2)(-u \cos \theta) + (M/2)v_2$
$u \cos \theta = -0.5 u \cos \theta + 0.5 v_2$
$1.5 u \cos \theta = 0.5 v_2$
$v_2 = 3 u \cos \theta$
The kinetic energy of the first part is $E_1 = \frac{1}{2} (M/2) (u \cos \theta)^2 = \frac{1}{4} M u^2 \cos^2 \theta$.
The kinetic energy of the second part is $E_2 = \frac{1}{2} (M/2) (3 u \cos \theta)^2 = \frac{1}{2} (M/2) (9 u^2 \cos^2 \theta) = \frac{9}{4} M u^2 \cos^2 \theta$.
Comparing $E_1$ and $E_2$:
$E_2 = 9 \times (\frac{1}{4} M u^2 \cos^2 \theta) = 9 E_1$.

(B) Let the centre of the bigger disc of radius $2R$ be at the origin $(0,0)$.
Let $M$ be the mass of the bigger disc. The mass per unit area is $\sigma = \frac{M}{\pi(2R)^2} = \frac{M}{4\pi R^2}$.
The mass of the removed smaller disc of radius $R$ is $m = \sigma \cdot \pi R^2 = \frac{M}{4\pi R^2} \cdot \pi R^2 = \frac{M}{4}$.
The centre of mass of the bigger disc is at $x_1 = 0$. The centre of mass of the removed smaller disc is at $x_2 = R$ (since the circumferences touch).
The centre of mass of the remaining part is given by:
$x_{CM} = \frac{M x_1 - m x_2}{M - m}$
$x_{CM} = \frac{M(0) - (M/4)(R)}{M - M/4} = \frac{-MR/4}{3M/4} = -\frac{R}{3}$.
The distance from the centre of the bigger disc is $|x_{CM}| = \frac{R}{3}$.
Comparing this with $\alpha R$,we get $\alpha = \frac{1}{3}$.

(B) The variation of the viscosity coefficient $(\eta)$ of liquids with temperature $(T)$ is described by the Andrade equation,which is given by the expression: $\eta = A e^{E / R T}$.
Here,$A$ is a constant,$E$ is the activation energy for viscous flow,$R$ is the universal gas constant,and $T$ is the absolute temperature.
As temperature increases,the viscosity of liquids decreases,which is consistent with the exponential form $\eta = A e^{E / R T}$.

(B) Given equations are $l+m+n=0$ and $2lm+2ln-mn=0$.
From the first equation,$l = -(m+n)$.
Substituting this into the second equation:
$2(-m-n)m + 2(-m-n)n - mn = 0$
$-2m^2 - 2mn - 2mn - 2n^2 - mn = 0$
$-2m^2 - 5mn - 2n^2 = 0$
$2m^2 + 5mn + 2n^2 = 0$
$(2m+n)(m+2n) = 0$
Case $1$: $2m+n=0 \Rightarrow m = -n/2$. Then $l = -(-n/2 + n) = -n/2$. The direction ratios are $\langle -n/2, -n/2, n \rangle$,which is proportional to $\langle -1, -1, 2 \rangle$.
Case $2$: $m+2n=0 \Rightarrow m = -2n$. Then $l = -(-2n + n) = n$. The direction ratios are $\langle n, -2n, n \rangle$,which is proportional to $\langle 1, -2, 1 \rangle$.
Let the direction ratios be $\vec{a} = \langle -1, -1, 2 \rangle$ and $\vec{b} = \langle 1, -2, 1 \rangle$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\vec{a} \cdot \vec{b} = (-1)(1) + (-1)(-2) + (2)(1) = -1 + 2 + 2 = 3$.
$|\vec{a}| = \sqrt{(-1)^2 + (-1)^2 + 2^2} = \sqrt{6}$.
$|\vec{b}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.
$\cos \theta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.
Therefore,$\theta = \cos^{-1}(1/2) = \frac{\pi}{3}$.