The volume of an ideal gas (\( \gamma = 1.5 \)) is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is:
- \( \frac{4}{5} \)
- \( \frac{16}{25} \)
- \( \frac{8}{5\sqrt{5}} \)
- \( \frac{2}{\sqrt{5}} \)
The Correct Option is C
Approach Solution - 1
To solve this problem, we need to use the relationship for adiabatic processes in ideal gases, given by the equation:
\(P_1 V_1^\gamma = P_2 V_2^\gamma\)
where:
- \(P_1\) is the initial pressure.
- \(V_1\) is the initial volume.
- \(P_2\) is the final pressure.
- \(V_2\) is the final volume.
- \(\gamma\) is the adiabatic index (given here as 1.5).
We can rearrange this equation to find the ratio of initial pressure to final pressure:
\(\frac{P_1}{P_2} = \left(\frac{V_2}{V_1}\right)^\gamma\)
Substituting the given values \(V_1 = 5\, \text{litres}\) and \(V_2 = 4\, \text{litres}\):
\(\frac{P_1}{P_2} = \left(\frac{4}{5}\right)^{1.5}\)
Calculating \(\left(\frac{4}{5}\right)^{1.5}\):
\(\left(\frac{4}{5}\right)^{1.5} = \left(\frac{4}{5}\right) \times \sqrt{\left(\frac{4}{5}\right)} = \frac{4}{5} \times \frac{2}{\sqrt{5}} = \frac{8}{5\sqrt{5}}\)
Thus, the ratio of initial pressure to final pressure is:
\(\frac{8}{5\sqrt{5}}\)
Hence, the correct answer is \(\frac{8}{5\sqrt{5}}\).
Approach Solution -2
For an adiabatic process, the relation between pressure and volume is:
\[ P_iV_i^\gamma = P_fV_f^\gamma, \] where \(\gamma = 1.5\).
Substitute the given volumes (\(V_i = 5 \, \text{litres}, V_f = 4 \, \text{litres}\)):
\[ P_i (5)^{1.5} = P_f (4)^{1.5}. \]
Rearranging for \(\frac{P_i}{P_f}\):
\[ \frac{P_i}{P_f} = \frac{(4)^{1.5}}{(5)^{1.5}}. \]
Simplify:
\[ \frac{P_i}{P_f} = \left(\frac{4}{5}\right)^{1.5}. \]
Write \(1.5\) as \(\frac{3}{2}\):
\[ \frac{P_i}{P_f} = \left(\frac{4}{5}\right)^{\frac{3}{2}} = \left(\frac{4}{5}\right)^1 \times \left(\frac{4}{5}\right)^{\frac{1}{2}}. \]
Simplify each term:
\[ \frac{P_i}{P_f} = \frac{4}{5} \times \sqrt{\frac{4}{5}} = \frac{4}{5} \times \frac{2}{\sqrt{5}} = \frac{8}{5\sqrt{5}}. \]
Final Answer: \(\frac{8}{5\sqrt{5}}\).
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Adiabatic Processes Questions
- \(\text{If three moles of monoatomic gas } \left( \gamma = \frac{5}{3} \right) \text{ is mixed with two moles of a diatomic gas } \left( \gamma = \frac{7}{5} \right),\)\(\text{the value of adiabatic exponent } \gamma \text{ for the mixture is:}\)
- A vessel contains \(0.15\,\text{m}^3\) of a gas at pressure \(8\) bar and temperature \(140^\circ\text{C}\) with \(c_p=3R\) and \(c_v=2R\). It expands adiabatically till pressure falls to \(1\) bar. The work done during this process is ____ kJ. (R is gas constant)}
- A cylinder with adiabatic walls is closed at both ends and is divided into two compartments by a frictionless adiabatic piston. Ideal gas is filled in both (left and right) the compartments at same \(P,V,T\). Heating is started from left side until pressure changes to \( \frac{27P}{8} \). If initial volume of each compartment was \(9\) litres then the final volume in right-hand side compartment is _____ litres. (for this ideal gas \(C_P/C_V=1.5\)).
- Initial pressure and volume of a monoatomic ideal gas are \(P\) and \(V\). The change in internal energy of this gas in adiabatic expansion to volume \(V_{\text{final}} = 8V\) is ________ J.
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.