First slide

Thermodynamic processes

Question

An ideal gas undergoes a circular cycle as shown in the figure. Find the ratio of maximum temperature of cycle to minimum temperature of cycle:

Difficult

A

${(\frac{1 + \sqrt{2}}{\sqrt{2} - 1})}^{2}$
B

${(\frac{2 + \sqrt{2}}{2 - \sqrt{2}})}^{2}$
C

${(\frac{3 + \sqrt{2}}{3 - \sqrt{2}})}^{2}$
D

${(\frac{4 + \sqrt{2}}{4 - \sqrt{2}})}^{2}$

Solution

The maximum temperature will occur at point A and minimum temperature will occur at point B of the cycle.
From symmetry we can say that coordinates of center of circle is $(2 P_{0}, 2 V_{0})$ and radius is $P_{0} o r V_{0}$ .
$∴ P_{A} = O C + O D = 2 P_{0} + (P_{0}) \cos 45^{0}$
$a n d P_{B} = O C - O E = 2 P_{0} - (P_{0}) \cos 45^{0}$
So, at point A
$\frac{P_{A}}{P_{0}} = \frac{V_{A}}{V_{0}} = 2 + c o s 45^{0} = = 2 + \frac{1}{\sqrt{2}} = \frac{2 \sqrt{2} + 1}{\sqrt{2}} = \frac{4 + \sqrt{2}}{2} \Rightarrow P_{A} = (\frac{4 + \sqrt{2}}{2}) P_{0} a n d V_{A} = (\frac{4 + \sqrt{2}}{2}) V$
${nRT}_{A} = P_{A} V_{A} = (\frac{4 + \sqrt{2}}{2}) P_{0} (\frac{4 + \sqrt{2}}{2}) V_{0} = P_{A} = {(\frac{4 + \sqrt{2}}{2})}^{2} P_{0} V_{0} - - - (1)$
Similarly at point B
$\frac{P_{B}}{P_{0}} = \frac{V_{B}}{V_{0}} = 2 - \cos 45^{0} = 2 - \frac{1}{\sqrt{2}} = \frac{2 \sqrt{2} - 1}{\sqrt{2}} = \frac{4 - \sqrt{2}}{2}$
${nRT}_{B} = P_{B} V_{B} = {(\frac{4 + \sqrt{2}}{2})}^{2} P_{0} V_{0} - - - - - (2) (1) / (2) w e g e t \Rightarrow \frac{T_{A}}{T_{B}} = {(\frac{4 + \sqrt{2}}{4 - \sqrt{2}})}^{2}$

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