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One mole of a monoatomic gas is brought from state 'A' to state 'B' along path ACB. Temperature at 'A' is 'To' Heat absorbed along path ACB is equal To

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$\frac{7}{2} R T_{0}$

$\frac{5}{2} R T_{0}$

$\frac{9}{2} R T_{0}$

$\frac{11}{2} R T_{0}$

answer is A.

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Detailed Solution

For AC

$\frac{P_{0} V_{0}}{T_{0}} = \frac{P_{0} (2 V_{0})}{T_{C}} \Rightarrow T_{C} = 2 T_{0}$

For CB

$\frac{P_{0}}{2 T_{0}} = \frac{2 P_{0}}{T_{B}} \Rightarrow T_{B} = 4 T_{0}$

For path ACB

$H = n C_{P} (T_{C} - T_{A})$

$+ n C_{V} (T_{B} - T_{C})$

$= n \frac{5}{2} R (T_{0}) + n \frac{3}{2} R (2 T_{0}) = \frac{11}{2} n R T_{0}$

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