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
+3 votes
88.0k views
in Physics by (60.8k points)

A gas is taken through a cyclic process ABC A as shown in figure (26-E6). If 2.4 cal of heat is given in the process, what is the value of J?

1 Answer

+3 votes
by (150k points)
edited by
 
Best answer

dQ = 2.4cal = 2.4J Joules
dw = WAB + WBC + WAC
= 0 + (1/2) × (100 + 200) × 103 200 × 10–6 – 100 × 103 × 200 × 10–6
= (1/2) × 300 × 103 200 × 10–6 – 20 = 30 – 20 = 10joules.
du = 0 (in a cyclic process)
dQ = dU +dW
=> 2.4 J = 10
=> J = 10/2.4 ≈ 4.17 J/Cal.

by (10 points)
+1
What is done in the last step??
by (10 points)
+1
2.4cal to 2.4 j without multiplying with 4.2
by (7.2k points)
+1
@anamika in last step the value of J is obtained through the equation 2.4 J = 10.
by (7.2k points)
+1
@sainikhil it's 2.4 cal = 2.4J joule
by (10 points)
+1
How did you find the work done in AB
by (150k points)
+1
for line AB
Change in volume, ΔV=0
So, W_(AB)=0

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

...