One mole of an ideal gas in inital state $A$ undergoes a cyclic process $A B C A$ , as shown in figure. Its pressure at $A$ is $P_{0}$ . Choose the correct option(s) from the following :

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

internal energies at $A$ and $B$ are the same

work done by the gas in process $A B$ is $P_{\circ} V_{\circ} l n 4$

pressure at $C$ is $\frac{P_{0}}{4}$

temperature at $C$ is $\frac{T_{0}}{4}$

answer is A, B, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Internal energy of an ideal gas depends on temperature . Since $T_{A} = T_{B} = T_{0}, w e c a n s a y U_{A} = U_{B} .$

$W_{A B} = n R T l n \frac{V_{2}}{V_{1}} = (1) (R) \frac{P_{0} V_{0}}{R} l n \frac{4 V_{0}}{V_{0}}$ = $P_{0} V_{0} \ln 4$

Since slope of the line BC is constant and it passes through the origin , process BC is isobaric process .So $P_{C} = P_{B} = \frac{P_{0} V_{0}}{4 V_{0}} = \frac{P_{0}}{4}$ .Also $\frac{\frac{P_{0}}{4} V_{0}}{T_{C}} = \frac{P_{0} V_{0}}{T_{0}} \Rightarrow T_{C} = \frac{T_{0}}{4}$