The given figure represents two isobaric processes for the same mass of an ideal gas, then [figure]
- P2>P1
- P1>P2
- P1=P2
- P2 P1
Answer
(2) P1>P2
Solution
PV=nRT V= ( nRP )T
isobaric V–T graph is a straight line through O with slope nRP 1P (same n)
the P2 line is steeper than the P1 line P2<P1
P1>P2
Answer: (2)