First slide

Heat ;work and internal energy

Question

One mole of an ideal monatomic gas has initial temperation T₀, is made to go through the cycle abca shown in the given figure. If U denotes the internal energy, then choose the correct alternative.

Moderate

A

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

$U_{c} - U_{b} = 3 {RT}_{0}$
C

$U_{c} - U_{a} = \frac{9 {RT}_{0}}{2}$
D

$U_{b} - U_{a} = \frac{3 {RT}_{0}}{2}$

Solution

One mole of an ideal monatomic gas (initial temperature T₀) is made to go through the cycle abca shown in figure. U denotes the internal energy.
For the process ab, $\frac{P_{0} V_{0}}{T_{0}} = \frac{2 P_{0} V_{0}}{T_{0}}$
$\begin{array}{l} T_{b} = 2 T_{0} \\ T_{b} > T_{a} \Rightarrow U_{b} > U_{a} \\ U_{b} - U_{a} = C_{v} ΔT = \frac{3 R}{2} (2 T_{0} - T_{0}) = \frac{3 {RT}_{0}}{2} \end{array}$
For the process bc,
$\begin{array}{l} \frac{P_{0} (2 V_{0})}{2 T_{0}} = \frac{2 P_{0} (2 V_{0})}{T_{C}} \Rightarrow T_{C} = 4 T_{0} \\ T_{c} > T_{0} \\ U_{c} - U_{b} = \frac{3 R}{2} (4 T_{0} - 2 T_{0}) = 3 {RT}_{0} \end{array}$
For the process ca,
$U_{c} - U_{a} = \frac{3 R}{2} (4 T_{0} - T_{0}) = \frac{9 {RT}_{0}}{2}$

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