Thermodynamics (Question series) Lesson-4 GATE Presented by Deep sangeet Maity

About ME Working as a Design engineer at "Cameron Manufacturing India Limited (A Schlumberger Company)". .Currently working as "GATE content editor and developer"at "Kreatryx" .Qualified GATE "3" times .Qualified "Gate 2018 with 99.4 percentile 779(75.26 marks) gate score" in mechanical stream .Qualified "Gate 2017 with 98.83 percentile 726(70 marks) gate score" irn mechanical stream I also have an 2+ years of experience of teaching Engineering subjects like Engineering mechanies, Thermodynamies, SOM, HMIT Ex-CADD Engineer at "CADD CENTRE". Passed B.E. with honours with 76.71%. I am pursuing my MTech in Thermal I have an 2.5+ years of experience as a CADD Engineer. . . also have worked for "CAD ACADEMY" as a Design engineer for 1 year.

The vapour pressure of liquid ammonia (in atmosphere) in the vicinity of the triple point can be expressed as 3063-15.16 where temperature T is expressed in K In a similar manner, the vapour pressure of solid ammonia can be expressed as In p +--=18.7 Take the molecular mass of ammonia to be 17 kg/kmol Q.1 The temperature and pressure at the triple point are (B) 295.2 K, 0.59 atm (D) 195.2 K, 0.59 atm (A) 295.2 K, 0.69 atm (C) 195.2 K, 0.69 atm The latent heat of vapourization is (A) 1298 kJ/kg Q.2 (B) 1398 kJ/kg (C) 1498 kJ/kg (D) 1698 kJ/kg

napon treasure of iquid am moma(in ahwp ammoan Imp +2759 = 17.0 A+ triple po =>IS. 16-2063 13-7-37S4 = 3754-3063-18.7-15, '

Tempenaturnan also poin+ et tiple of otid & id ammonia 30 63 375 1 37 S -1 Lia . ammorm 3o 6 3

04 .in p +3063 15. K -3063 0. cl

3063P 30 63PT -h 3063 8.315 #3-149 8 k 31 P . er 63

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Deep Sangeet Maity

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Puja Kumari

7 months ago

Hello sir
I solve this question by method written below..
please let me know if I am doing right..
25 32 48 82 x
32-25=7
48-32=16
82-48=34
7 16 34 I have taken it as (7+1)*2= 16 (16+1)*2=34 and (34+1)*2=70
so value of x = 82+70 i.e. 152

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Anuraj MAURYA

3 months ago

Why we treated vfg as volume

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Deep Sangeet Maity

3 months ago

Becoz it is a phase change

Suraj prasad singh

6 months ago

sir a condition is given that ∆Q/T =0 and ∆S greater than 0 and in options it is given that is it reversible adiabatic or irreversible adiabatic

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Deep Sangeet Maity

6 months ago

it's a contradictory Question.... is there any other options?

Deep Sangeet Maity

6 months ago

see if ∆S(system or surrounding)>0 and ∆Q/T=0 then it may or may not be reversible.... if Q is zero then it is a adiabatic process.... if ∆s(universe)>0 then it is irreversible...