$\frac{3}{25}$ mole of an ideal monatomic gas undergoes a process as shown in figure. If temperature at point A is 100K then which of the following option(s) is (are) correct. $(t a k e R = \frac{25}{3} S I u n i t)$

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Change in internal energy in process AB is 450 J

Work done in cycle process ABCDA 362.5 J

Work done in process AB is 150 J

Change in internal energy in process CD is 1050 J

answer is A, B, C, D.

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Detailed Solution

$\frac{1}{2} (P_{0} + 2 P_{0}) 4 V_{0} = 6 P_{0} V_{0}$

For path AB,

$P \propto V T \propto V^{2} \frac{T_{A}}{T_{B}} = {(\frac{4 V_{0}}{8 V_{0}})}^{2} T_{B} = 400 K Δ U_{A B} = n C_{V} Δ T - \frac{3}{25} \frac{3}{2} (\frac{25}{5}) \times 300 = 450 J W_{CYCLE} = - (4 P_{0}) 7 V_{0} + \frac{1}{2} (R_{0} + 4 R_{0}) 3 V_{0} + P_{0} V_{0} = (28 + \frac{15}{2} + 1) P_{0} V_{0} = \frac{- 39}{2} P_{0} V_{0}$

$Δ U_{C D} = n C_{V} Δ T = \frac{3}{25} \frac{3}{2} (- 700) T_{B} = 400 L T_{C} = 800 K T_{B} = 100 K \begin{array}{l} F o r C D \\ V \propto T \end{array}$