An ideal gas undergoes a process maintaining relation between pressure (P) and volume (V) as P=P0 (1+ ( V0V )2 )-1, where P0 and V0 are constants. If two samples A and B (two moles each) with initial volumes V0 and 3V0 respectively undergo above mentioned process and attain same pressure, then the difference in the temperatures of these samples, TB-TA is . (R = gas constant)
- 9P0V08R
- 11P0V010R
- 7P0V06R
- 13P0V011R
Answer
(2) 11P0V010R
Solution
Process: P=P0 (1+ V02V2 )-1= P0V2V2+V02
Temperature of each sample (n = 2): T= PVnR= PV2R
Sample A, V=V0: PA= P02 ⇒ TA= (P0/2)V02R= P0V04R= 5P0V020R
Sample B, V=3V0: PB= 9P010 ⇒ TB= (9P0/10)(3V0)2R= 27P0V020R
∴ TB-TA= 27P0V0-5P0V020R= 11P0V010R
Answer: (2)