One mole of ideal monatomic gas is taken through the cyclic process ABCDA as shown in figure. Match the Quantities in column I with that of column II
Column-I | Column-II |
(i) work done in process BC | [A] – 3/2 RT0 |
(ii) Change in internal energy in CD | [B] 3/2 RT0 |
(iii) Heat transferred in in process DA | [C] – 5RT0 |
(iv) Heat transferred in in process AB | [D] 2 RT0 |
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Ans.
(i) [D]
(ii) [A]
(iii) [C]
(iv) [B]
Sol.
AB: isochoric process
T B = 2T 0
work done in BC: (isobaric)
W = P Δ V = n R Δ T
W = nR (4T 0 –2T 0 )
W = (1) 2RT 0
W BC = 2RT 0
Internal energy change in CD
CD: Δ U CD = nC v (T D –T C )
= (1) 
R (3T 0 –4T 0 )
= – 
RT 0
Heat transferred in DA: isobaric
Q = nCp Δ T
= (1) 
(T 0 –3T 0 )
= 
× (–2T 0 )
= – 5RT 0
Heat transferred in AB (isochoric)
Q = Δ U + W
W = 0
Q = Δ U = nC V (T B – T A )
Q = (1) 
R (2T 0 – T 0 )
Q = 
RT 0
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems