1

### JEE Advanced 2021 Paper 2 Online

Numerical
One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are same, the value of $$\ln {{{V_3}} \over {{V_2}}}$$ is _________.

(U : internal energy, S : entropy, p : pressure, V : volume, R : gas constant)

(Given : molar heat capacity at constant volume, CV,m of the gas is $${5 \over 2}$$R)

## Explanation

For process 1, entropy is constant thus, q is constant

$$\therefore$$ $${W_I} = \Delta U = n{C_{V,m}}\Delta T$$

$$= - (2250 - 450)R = - 1800R$$ .... (i)

$$\Delta U = n{C_{V,m}}\Delta T$$

$$- 1800R = 1 \times {{5R} \over 2} \times \Delta T$$

$$\Rightarrow \Delta T = - 720K$$

$${T_2} - {T_1} = - 720K$$

$${T_2} = - 720K + 900K = 180K$$

For process II,

$${W_{II}} = - nR{T_2}\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$$

$$= - 1 \times R \times 180\ln \left( {{{{V_3}} \over {{V_2}}}} \right)$$ .... (ii)

As, both work done for process I and II are equal,

Therefore, WI = WII

$$- 1800R = - R \times 180 \times \ln \left( {{{{V_3}} \over {{V_2}}}} \right)$$

$$\ln \left( {{{{V_3}} \over {{V_2}}}} \right) = {{1800} \over {180}} \Rightarrow \ln \left( {{{{V_3}} \over {{V_2}}}} \right) = 10$$
2

### JEE Advanced 2021 Paper 1 Online

Numerical
For the reaction, X(s) $$\rightleftharpoons$$ Y(s) + Z(g), the plot of $$\ln {{pz} \over {{p^\theta }}}$$ versus $${{{{10}^4}} \over T}$$ is given below (in solid line), where pz is the pressure (in bar) of the gas Z at temperature T and $${{p^\theta }}$$ = 1 bar.

(Given, $${{d(\ln K)} \over {d\left( {{1 \over T}} \right)}} = - {{\Delta {H^\theta }} \over R}$$, where the equilibrium
constant, $$K = {{pz} \over {{p^\theta }}}$$ and the gas constant, R = 8.314 J K$$-$$1 mol$$-$$1)

The value of $$\Delta$$S$$\theta$$ (in J K$$-$$1 mol$$-$$1) for the given reaction, at 1000 K is _________.

3

### JEE Advanced 2021 Paper 1 Online

Numerical
For the reaction, X(s) $$\rightleftharpoons$$ Y(s) + Z(g), the plot of $$\ln {{pz} \over {{p^\theta }}}$$ versus $${{{{10}^4}} \over T}$$ is given below (in solid line), where pz is the pressure (in bar) of the gas Z at temperature T and $${{p^\theta }}$$ = 1 bar.

(Given, $${{d(\ln K)} \over {d\left( {{1 \over T}} \right)}} = - {{\Delta {H^\theta }} \over R}$$, where the equilibrium
constant, $$K = {{pz} \over {{p^\theta }}}$$ and the gas constant, R = 8.314 J K$$-$$1 mol$$-$$1)

The value of standard enthalpy, $$\Delta$$H$$\theta$$ (in kJ mol$$-$$1) for the given reaction is _______.

4

### JEE Advanced 2020 Paper 1 Offline

Numerical
Aluminium reacts with sulphuric acid to form aluminium sulphate and hydrogen. What is the volume of hydrogen gas in litre (L) produced at 300 K and 1.0 atm pressure, when 5.4 g of aluminium and 50.0 mL of 5.0 M sulphuric acid are combined for the reaction?

(Use molar mass of aluminium as 27.0 g mol$$-$$1, R = 0.082 atm L mol$$-$$1 K$$-$$1)

## Explanation

Aluminium reacts with sulphuric acid to form aluminium sulphate and hydrogen.

$$\mathop {2Al}\limits_{\left( {{{5.4} \over {27}} = 0.2\,mol} \right)} + \mathop {3{H_2}S{O_4}}\limits_{\left( {{{50 \times 5} \over {1000}} = 0.25\,mol} \right)} \buildrel {} \over \longrightarrow A{l_2}{(S{O_4})_3} + 3{H_2}$$

H2SO4 is limiting reagent and moles of H2(g) produced = 0.25 mol

Using ideal gas equation,

pV = nRT

$$\Rightarrow$$ $$V = {{0.25 \times 0.082 \times 300} \over {1\,atm}} = 6.15\,L$$

On those following papers in Numerical
Number in Brackets after Paper Indicates No. of Questions
JEE Advanced 2021 Paper 2 Online (1)
JEE Advanced 2021 Paper 1 Online (2)
JEE Advanced 2020 Paper 2 Offline (1)
JEE Advanced 2020 Paper 1 Offline (1)
JEE Advanced 2018 Paper 2 Offline (1)

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