1
JEE Advanced 2020 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1
If the distribution of molecular speeds of a gas is as per the figure shown below, then the ratio of the most probable, the average, and the root mean square speeds, respectively, is


2
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
The standard state Gibbs free energies of formation of $$C$$(graphite) and $$C$$(diamond) at $$T=298$$ $$K$$ are
$${\Delta _f}{G^0}$$ [$$C$$(graphite)] $$ = 0kJmo{l^{ - 1}}$$
$${\Delta _f}{G^0}$$ [$$C$$(diamond)] $$ = 2.9kJmo{l^{ - 1}}$$
The standard state means that the pressure should be $$1$$ bar, and substance should be pure at a given temperature. The conversion of graphite [$$C$$(graphite)] to diamond [$$C$$(diamond)] reduces its volume by $$2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}$$ If $$C$$(graphite) is converted to $$C$$(diamond) isothermally at $$T=298$$ $$K,$$ the pressure at which $$C$$(graphite) is in equilibrium with $$C$$(diamond), is
[Useful information : $$1$$ $$J=1$$ $$kg\,{m^2}{s^{ - 2}};1\,Pa = 1\,kg\,{m^{ - 1}}{s^{ - 2}};$$ $$1$$ bar $$ = {10^5}$$ $$Pa$$]
$${\Delta _f}{G^0}$$ [$$C$$(graphite)] $$ = 0kJmo{l^{ - 1}}$$
$${\Delta _f}{G^0}$$ [$$C$$(diamond)] $$ = 2.9kJmo{l^{ - 1}}$$
The standard state means that the pressure should be $$1$$ bar, and substance should be pure at a given temperature. The conversion of graphite [$$C$$(graphite)] to diamond [$$C$$(diamond)] reduces its volume by $$2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}$$ If $$C$$(graphite) is converted to $$C$$(diamond) isothermally at $$T=298$$ $$K,$$ the pressure at which $$C$$(graphite) is in equilibrium with $$C$$(diamond), is
[Useful information : $$1$$ $$J=1$$ $$kg\,{m^2}{s^{ - 2}};1\,Pa = 1\,kg\,{m^{ - 1}}{s^{ - 2}};$$ $$1$$ bar $$ = {10^5}$$ $$Pa$$]
3
JEE Advanced 2016 Paper 1 Offline
MCQ (Single Correct Answer)
+3
-1
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to
2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surrounding ($$\Delta$$Ssurr)in
JK–1 is (1L atm = 101.3 J)
4
JEE Advanced 2015 Paper 1 Offline
MCQ (Single Correct Answer)
+8
-4
Match the thermodynamics processes given under column I with expression given under column II
Column I
(A) Freezing water at 273 K and 1 atm
(B) Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions.
(C) Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container.
(D) Reversible heating of H2(g) at 1 atm from 300K to 600K, followed by reversible cooling to 300K at 1 atm
Column II
(p) q = 0
(q) w = 0
(r) $$\Delta S_{sys}$$ < 0
(s) $$\Delta U$$ = 0
(t) $$\Delta G$$ = 0
Column I
(A) Freezing water at 273 K and 1 atm
(B) Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions.
(C) Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container.
(D) Reversible heating of H2(g) at 1 atm from 300K to 600K, followed by reversible cooling to 300K at 1 atm
Column II
(p) q = 0
(q) w = 0
(r) $$\Delta S_{sys}$$ < 0
(s) $$\Delta U$$ = 0
(t) $$\Delta G$$ = 0
Questions Asked from Thermodynamics (MCQ (Single Correct Answer))
Number in Brackets after Paper Indicates No. of Questions
JEE Advanced 2020 Paper 1 Offline (1)
JEE Advanced 2017 Paper 2 Offline (1)
JEE Advanced 2016 Paper 1 Offline (1)
JEE Advanced 2015 Paper 1 Offline (1)
JEE Advanced 2014 Paper 2 Offline (1)
JEE Advanced 2013 Paper 2 Offline (2)
JEE Advanced 2013 Paper 1 Offline (1)
IIT-JEE 2012 Paper 2 Offline (1)
IIT-JEE 2011 Paper 2 Offline (1)
IIT-JEE 2010 Paper 1 Offline (1)
IIT-JEE 2008 Paper 2 Offline (1)
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