A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant pressure until volume is doubled. The average molar heat capacity for whole process is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{23 R}{6}$

$\frac{13 R}{6}$

$\frac{17 R}{6}$

$\frac{19 R}{6}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Isochoric process:

$\frac{P}{T} = \frac{2 P}{T_{2}} \Rightarrow T_{2} = 2 T$

Isobaric process

$\frac{V_{2}}{T_{2}} = \frac{V_{3}}{T_{3_{}}} \frac{V_{2}}{2 T_{1}} = \frac{2 V_{2}}{T_{3_{}}} \Rightarrow T_{3} = 4 T_{1} c h a n g e i n t e m e p e r a t u e = Δ T = 4 T_{1} - T_{1} = 3 T_{1} F O r d i a t o m i c C_{v} = \frac{5 R}{2} a n d C_{p} = \frac{7 R}{2} T_{1} = T_{0}, V_{1} = V_{0} a n d P_{1} = P_{0}$

Let initial pressure, volume, temperature be $P_{0}, V_{0}, T_{0}$ 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 $B (2 P_{0}, 2 V_{0}, 4 T_{0})$ isobarically