Correct Answer - B
b. Let initial pressure, volume and temperature be `P_(0), V_(0)` and `T_(0)`, repectively, indicated by state `A` in `P -V` diagram. The gas is then isochorically taken to state `B (2 P_(0), V_(0), 2 T_(0))` and then taken from state `B` to state `C (2 P_(0), 2 V_(0), 4 T_(0))` isobarically.
Total heat absorbed by 1 mol of gas
`Delta Q = C_(v) (2 T_(0) - T_(0)) + C_(P) (4 T_(0) - 2 T_(0))`
`= (5)/(2) RT_(0) + (7)/(2) R xx 2T_(0) = (19)/(2) RT_(0)`
Total change in temperature from series `A` to `C` is `Delta T = 3 T_(0)`
Therefore,
Molar heat capacity `= (Delta Q)/(Delta T) = ((19)/(2) RT_(0))/(3 T_(0)) = (19)/(6) R`