One mole of an ideal monatomic gas has initial temperature $T_{0},$ is made to go through the cycle $a b c a$ shown in figure.

If $U$ denotes the internal energy, then choose the correct alternative.

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$U_{c} > U_{b} > U_{a}$

$U_{c} - U_{b} = 3 R T_{0}$

$U_{c} - U_{a} = \frac{9 R T_{0}}{2}$

$U_{b} - U_{a} = \frac{3 R T_{0}}{2}$

answer is A, B, C, D.

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Detailed Solution

One mole of an ideal monatomic gas (initial temperature $T_{0}$ ) is made to go through the cycle $a b c a$ shown in figure $U$ denotes the internal energy.
For the process $a b, \frac{P_{0} V_{0}}{T_{0}} = \frac{2 P_{0} V_{0}}{T_{b}}$
$T_{b} = 2 T_{0}$
$T_{b} > T_{a} \Rightarrow U_{b} > U_{a}$
$U_{b} - U_{a} = C_{V} Δ T = \frac{3 R}{2} (2 T_{0} - T_{0}) = \frac{3 R T_{0}}{2}$
For the process $b c,$
$\frac{P_{0} (2 V_{0})}{2 T_{0}} = \frac{2 P_{0} (2 V_{0})}{T_{c}} \Rightarrow T_{c} = 4 T_{0}$
$T_{c} > T_{0}$
$U_{c} - U_{b} = \frac{3 R}{2} (4 T_{0} - 2 T_{0}) = 3 R T_{0}$
For the process $c a,$
$U_{c} - U_{a} = \frac{3 R}{2} (4 T_{0} - T_{0}) = \frac{9 R T_{0}}{2}$