(p - V) diagram of (n) moles of an ideal gas is as shown in figure. Find the maximum temperature between (A) and (B).
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How to proceed For given number of moles of a gas,
` T prop pV (p V = nRT)`
Although `(p V)_A = (p V)_B = 2p_0 V_0 or T_A = T_B`, yet it is not an isothermal process. Because in isothermal process (p- V) graph is a rectangular hyperbola while it is a straight line. So, to see the behaviour of temperature, First we will find either (T - V) equation or (T _ p) equation and from that equation we can judge how the temperature baries. From the graph, first we will write )p V) equation, then we will convert it either in (T - V) equation or in (T - p) quation with the help of equation, (pV = nRT).

Answer

A, D

Explanation

From the graph the (p - V) equation can be written as
`p = -((p_0)/(V_0))V + 3 P_0` `(y = -mx + c)`
or `pV = -((p_0)/(V_0))V^2 + 3 p_0 V`
or `n R T = 3 p_0 V -((p_0)/(V_0))V^2` ( as pV =nRT)
or `T = 1/(nR)[3 p _0V - ((p_0)/V_0)V^2]`
This is the required (T - V) equation. This is quadratic in (V). Hence, (T - V) graph is a parabola. now, to find maximum or minimum value of (T), we can substitute.
`(dT)/(dV) = 0`
or `3 p_0 -((2 p_0)/(V_0))V = 0 or V = 3/2 V_0`
Further `(d^2 T)/(dV^2)` is negative at ` V = 3/2 v_0`.
Hence, (t) is maximum at `v = 3/2 v_0` and this maximum value is
`T_(max) = 1/(n R)[(3 p_0)((3 V_0)/2)-((p_0)/(V_0))((3 V_0)/2)^2]`
or `T_(max) = (9 p_0 V_0)/(4 n R)`
Thus, (T - V) graph is as shown in figure.(##DCP_V03_C20_S01_031_S01##).
`T_A = T_B = (2 p_0 V_0)/(n R) and T_(max) =(9 p_0 V_0)/(4 n R)`
= `2.25 (p_0 V_0)/(n R)`.