JEE Advanced 2015Paper 1
- Exam
- JEE Advanced
- Year
- 2015
- Paper
- Paper 1
- Subject
- Chemistry
- Chapter
- Thermodynamics (C)
JEE Advanced 2015 · Paper 1 · Chemistry · Thermodynamics (C)
Match the thermodynamic processes given under Column I with the...
Match the thermodynamic processes given under Column I with the expressions given under Column II.
| Column I | Column II | ||
| A. | Freezing of water at 273 K and 1 atm | P. | |
| B. | Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions | Q. | |
| C. | Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container | R. | |
| D. | Reversible heating of at 1 atm from 300 K to 600 K, followed by reversible cooling to 300 K at 1 atm | S. | |
| T. |
Options
- Aa-r;b-s,t;c-r,s,t;d-s,t;
- Ba-q,r;b-q;c-q;d-s,t;
- Ca-r,t;b-p,q,s;c-p,q,s;d-p,q,s,t;
- Da-q,r,s;b-s,t;c-q,r,s;d-t;
Correct answer
C. a-r,t;b-p,q,s;c-p,q,s;d-p,q,s,t;
Step-by-step solution
A - at 273 K. & 1 atm
(as water expands on freezing),
B - Free expansion of ideal gas.
q = 0
w = 0
C - Mixing of equal volume of ideal gases at constant pressure & temp in an isolated container
q = 0, w = 0, , ,
D - 600K 300Kq = 0, w = 0, ∆ U = 0∆ G = 0∆ {S}_{sys} = 0$
(as water expands on freezing),
B - Free expansion of ideal gas.
q = 0
w = 0
C - Mixing of equal volume of ideal gases at constant pressure & temp in an isolated container
q = 0, w = 0, , ,
D - 600K 300Kq = 0, w = 0, ∆ U = 0∆ G = 0∆ {S}_{sys} = 0$