wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An ideal gas (molar specific heat CV = 5R2) is taken along paths acb, adb and ab. P2 = 2P1, V2 = 2V1. Along ab, P = kV where k is a constant. The various parameters are shown in figure. Match the column I with the corresponding option of column II.

Column I Column II
i. Wacb p 15RT12
ii. Wadb q −15RT12
iii. Uab r RT1
iv. Ubca s 2RT1

Now match the given columns and select the correct option from the codes given below.

A
i - r, ii - s, iii - p, iv - q
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
i - s, ii - r, iii - p, iv - q
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
i - r, ii - s, iii - q, iv - p
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
i - s, ii - r, iii - q, iv - p
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B i - s, ii - r, iii - p, iv - q
Wacb = Wac + Wcb = 0 + P2(V2 −V1) = P2V1 = 2P1V1 = 2RT1(∵ V2 = 2V1)Wadb = Wad + Wdb = P1(V2 − V1) + 0= P1V1 = RT1Uab = Uac + Ucb = (Qac − Wac) + (Qcb − Wcb) = CV(Tc − T1) + CV(T2 − TC)= CV (T2 −T1)= 5R2 (T2 −T1)(Given CV =5R2)For an ideal gas,P1V1T1 = P2V2T2P2 = 2P1 and V2 = 2V1⇒ T2 = 4T1∴Uab = 5R2 (T2 −T1) = 15RT12 For the entire cycle, ΔU = 0⇒Ubca = − Uab Ubca =−15RT12

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Carbon Monoxide
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon