An ideal gas undergoes a circular cycle cente | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) From the given cyclic process,the equation of the circle is:$(p-4)^2 + (V-4)^2 = 2^2 = 4 \quad \dots(i)$From the ideal gas equation,$pV = nRT$. Fo

Vedclass · Accepted Answer

$\frac{30}{R}$

Vedclass · Answer

(A) From the given cyclic process,the equation of the circle is:$(p-4)^2 + (V-4)^2 = 2^2 = 4 \quad \dots(i)$From the ideal gas equation,$pV = nRT$. For $n=1 \, mol$,$T = \frac{pV}{R}$.To find the maximum temperature,we need to maximize the product $pV$.Let $y = pV$. From $(i)$,$p = 4 \pm \sqrt{4 - (V-4)^2}$.Substituting this into $y = pV$,we get $y = V(4 \pm \sqrt{4 - (V-4)^2})$.To maximize $y$,we set $\frac{dy}{dV} = 0$. Alternatively,using the property of a circle,the maximum value of $pV$ occurs at the point on the circle where the tangent is perpendicular to the line connecting the origin $(0,0)$ to the point $(p,V)$.The line from the origin to the center $(4,4)$ has a slope of $1$. The tangent at the maximum $pV$ point must have a slope of $-1$.The coordinates of the point on the circle $(p-4)^2 + (V-4)^2 = 2^2$ with tangent slope $-1$ are given by:$p = 4 + 2 \cos(135^\circ) = 4 - \sqrt{2}$$V = 4 + 2 \sin(135^\circ) = 4 + \sqrt{2}$Wait,for maximum $pV$,we look for the point furthest from the origin. The distance from the origin to the center is $\sqrt{4^2 + 4^2} = 4\sqrt{2}$. The radius is $2$. The maximum distance from the origin to the circle is $4\sqrt{2} + 2$.The point $(p,V)$ on the circle that maximizes $pV$ is $p = 4 + \sqrt{2}$ and $V = 4 + \sqrt{2}$.Thus,$(pV)_{\max} = (4+\sqrt{2})(4+\sqrt{2}) = 16 + 2 + 8\sqrt{2} = 18 + 8(1.414) = 18 + 11.312 = 29.312$.Therefore,$T_{\max} = \frac{(pV)_{\max}}{R} \approx \frac{29.3}{R} \approx \frac{30}{R}$.

An ideal gas undergoes a circular cycle centered at $4 \, atm, 4 \, L$ as shown in the diagram. The maximum temperature attained in this process is close to

Similar Questions

In the given figure,$1$ represents isobaric,$2$ represents isothermal,and $3$ represents adiabatic processes of an ideal gas. If $\Delta U_{1}, \Delta U_{2}, \Delta U_{3}$ are the changes in internal energy in these processes respectively,then:

An ideal monoatomic gas is taken round the cycle $12341$ shown in the figure ($p-V$ diagram). The work done in one cycle is (in $P_0V_0$)?

Using the given $P-V$ diagram, the work done by an ideal gas along the path $\text{ABCD}$ is $-$ (in $\text{P}_0 \text{V}_0$)

$A$ monoatomic ideal gas goes through a cyclic process as shown in the figure. The efficiency of this process is (in $\%$)

One mole of an ideal diatomic gas undergoes a transition from $A$ to $B$ along a path $AB$ as shown in the figure. The change in internal energy of the gas during the transition is

Vedclass Test Series

Exam Paper Generator

Online Exam Module