Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
7.7k views
in Chemical thermodynamics by (64.8k points)

The vapor pressure of H2O (l) at 353 K is 532 mm Hg. The external pressure on H2O (l) taken in a cylinder fitted with frictionless movable piston initially containing 0.9 L (=0.9 kg) of H2O (l) at 33 K is increased to 1 atm. Temperature remained constant. Now, heat is supplied keeping pressure constant till 0.45 L of H2O (l) (=0.45 kg) is evaporated to form H2O (g) at 373 K. carefully observe the diagrams provided and form given data, answer the following questions Given:

Specific heat of H2O = 4.2J/gm⁰C

Use Hvap at 373 K and 1 atm =+40 kJ/mol

1L atm = 100 Joule

1 atm = 760 mm Hg

R = 8 Joule/mole K 

(Assume internal energy of liquid to be dependent only on temperature).

 What is the work done in going from state 1 to state 3 in Joules? 

(A) Zero 

(B) 74.6 

(C) 90 

(D) 31.5

Please log in or register to answer this question.

1 Answer

+1 vote
by (58.4k points)

Correct option (B) 74.6

Explanation:

Work done in 1 to 3

w1–2 = 0 as ∆U = 0

w2–3 = Pext dU = (0.45 x 103/18 x 0.0821 x 373/100)

w2 (10 KJ) 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...