An ideal gas initially at 0°C temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is \( \frac{3}{2} \), the change in temperature due to the thermodynamics process is K.
An ideal gas initially at 0°C temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is \( \frac{3}{2} \), the change in temperature due to the thermodynamics process is K.
Show Hint
Approach Solution - 1
Given Data:
Initial temperature, $T_1 = 0^\circ C = 273 \text{ K}$
Final volume, $V_2 = \dfrac{V_1}{4}$
Ratio of specific heats, $\gamma = \dfrac{C_p}{C_v} = \dfrac{3}{2}$
Step 1: Relation for adiabatic process
For a sudden (adiabatic) compression, $T_2 V_2^{\gamma - 1} = T_1 V_1^{\gamma - 1}$
Step 2: Substitute given values
$\displaystyle \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1}$
$\displaystyle \frac{T_2}{273} = \left(\frac{V_1}{V_1/4}\right)^{\frac{3}{2} - 1}$
$\displaystyle \frac{T_2}{273} = 4^{\frac{1}{2}} = 2$
Step 3: Calculate final temperature
$T_2 = 2 \times 273 = 546 \text{ K}$
Step 4: Change in temperature
$\Delta T = T_2 - T_1 = 546 - 273 = 273 \text{ K}$
Final Answer:
$\boxed{273\ \text{K}}$
Approach Solution -2
To determine the temperature change during an adiabatic compression of an ideal gas, we follow these steps:
1. Initial Conditions:
- Initial temperature (T₁) = 0°C = 273 K
- Initial volume (V₁)
- Final volume (V₂) = V₁/4
- Ratio of specific heats (γ) = Cₚ/Cᵥ = 3/2
2. Adiabatic Process Relation:
For an adiabatic process, we use:
\[
T V^{\gamma - 1} = \text{constant}
\]
Which gives the relation:
\[
T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}
\]
3. Calculate Final Temperature:
Solving for the final temperature T₂:
\[
T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} = 273 \left( \frac{V_1}{V_1/4} \right)^{1/2} = 273 \times (4)^{1/2} = 273 \times 2 = 546 \text{ K}
\]
4. Determine Temperature Change:
The change in temperature is:
\[
\Delta T = T_2 - T_1 = 546 \text{ K} - 273 \text{ K} = 273 \text{ K}
\]
Key Points:
- The process is adiabatic (no heat transfer)
- Volume reduces to 1/4th of original
- Temperature doubles due to the compression
- γ = 3/2 indicates a monatomic ideal gas
Final Answer:
The temperature increases by \(\boxed{273 \text{ K}}\).
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 Thermodynamics and Work Done Questions
Match the List-I with List-II
Choose the correct answer from the options given below:A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J, then the mass of the bullet is grams. Given Data: Latent heat of fusion of lead = \(2.5 \times 10^4 \, \text{J kg}^{-1}\) and specific heat capacity of lead = 125 J kg\(^{-1}\) K\(^{-1}\).
Using the given P-V diagram, the work done by an ideal gas along the path ABCD is:
- For a diatomic gas, if \( \gamma_1 = \frac{C_P}{C_V} \) for rigid molecules and \( \gamma_2 = \frac{C_P}{C_V} \) for another diatomic molecules, but also having vibrational modes. Then, which one of the following options is correct? (where \( C_P \) and \( C_V \) are specific heats of the gas at constant pressure and volume)
- Find out magnitude of work done in the process ABCD (in kJ).
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 ______.