1
JEE Advanced 2021 Paper 2 Online
Numerical
+4
-0
Change Language
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 _________.

JEE Advanced 2021 Paper 2 Online Chemistry - Thermodynamics Question 15 English
(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)
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2
JEE Advanced 2021 Paper 1 Online
Numerical
+2
-0
Change Language
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.

JEE Advanced 2021 Paper 1 Online Chemistry - Thermodynamics Question 16 English Comprehension
(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, $$\Delta$$Ho (in kJ mol$$-$$1) for the given reaction is _______.
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3
JEE Advanced 2021 Paper 1 Online
Numerical
+2
-0
Change Language
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.

JEE Advanced 2021 Paper 1 Online Chemistry - Thermodynamics Question 18 English Comprehension
(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 _________.
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4
JEE Advanced 2020 Paper 2 Offline
Numerical
+4
-0
Change Language
Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K) at which the reduction of cassiterite by coke would take place.

At $$298K:{\Delta _f}H^\circ [Sn{O_2}(s)] = - 581.0$$ mol-1,

$$\eqalign{ & {\Delta _f}H^\circ [(C{O_2})(g)] = - 394.0\,kJ\,mol{ ^{-1}} \cr & S^\circ [Sn{O_2}(s)] = 56.0J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [Sn(s)] = 52.0\,J\,K{ ^{-1}}mo{l^{ - 1}} \cr & S^\circ [C(s)] = 6.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [C{O_2}(g)] = 210.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr} $$

Assume that, the enthalpies and the entropies are temperature independent.
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