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Thermodynamic Processes

One mole of a...

Question

One mole of an ideal monatomic gas has initial temperature $T_{0}$ , is made to go through the cycle $a b c a$ as shown in fig. If $U$ denotes the internal energy, then choose the correct alternative.

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

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

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

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

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Solution

The correct option is D $U_{b} - U_{a} = \frac{3 R T_{0}}{2}$

Ideal gas equation for one mole of an ideal monatomic gas is
$\frac{P V}{T} = constant$

For $a b$ process [isobaric]
$\frac{V_{a}}{T_{a}} = \frac{V_{b}}{T_{b}} \Rightarrow \frac{V_{0}}{T_{0}} = \frac{2 V_{0}}{T_{b}}$
$\Rightarrow T_{b} = 2 T_{0} \Rightarrow T_{b} > T_{a} \Rightarrow U_{b} > U_{a}$
( $∵$ Internal energy $U \propto T$ )
$U_{b} - U_{a} = C_{v} Δ T = \frac{3 R}{2} (2 T_{0} - T_{0}) = \frac{3 R T_{0}}{2}$

For process $b c$ (isochoric process)
$\frac{P_{b}}{T_{b}} = \frac{P_{c}}{T_{c}} \Rightarrow \frac{P_{0}}{2 T_{0}} = \frac{2 P_{0}}{T_{c}} \Rightarrow T_{c} = 4 T_{0}$
$\Rightarrow T_{c} > T_{a} \Rightarrow U_{c} > U_{a}$
$U_{c} - U_{b} = \frac{3 R}{2} (4 T_{0} - 2 T_{0}) = 3 R T_{0}$

For Process $c a$ :
$U_{c} - U_{a} = \frac{3 R}{2} (4 T_{0} - T_{0}) = \frac{9 R T_{0}}{2}$
Thus, $a, b, c, d$ are the correct

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