Question:
A gas at \(37^\circ\text{C}\) is compressed adiabatically to half of its volume. Then the final temperature of the gas is
\[
(\text{Ratio of specific heat capacities of the gas is }1.5)
\]
A gas at \(37^\circ\text{C}\) is compressed adiabatically to half of its volume. Then the final temperature of the gas is
\[ (\text{Ratio of specific heat capacities of the gas is }1.5) \]
\[ (\text{Ratio of specific heat capacities of the gas is }1.5) \]
Show Hint
For an adiabatic process,
\[
TV^{\gamma-1}=\text{constant}.
\]
If a gas is compressed, its temperature increases according to
\[
T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}.
\]
Always convert temperatures to Kelvin before applying thermodynamic relations.
Updated On: Jun 26, 2026
- \(165.3^\circ\text{C}\)
- \(438.3^\circ\text{C}\)
- \(400^\circ\text{C}\)
- \(0^\circ\text{C}\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Write the given data.
Initial temperature: \[ T_1=37^\circ\text{C}=310\ \text{K}. \] Ratio of specific heats: \[ \gamma=1.5=\frac{3}{2}. \] The gas is compressed to half its original volume, therefore \[ V_2=\frac{V_1}{2}. \] Hence, \[ \frac{V_1}{V_2}=2. \]
Step 2: Use the adiabatic relation.
For an adiabatic process, \[ TV^{\gamma-1}=\text{constant}. \] Therefore, \[ T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}. \] Hence, \[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1}. \]
Step 3: Substitute the given values.
Since \[ \gamma-1 = \frac{3}{2}-1 = \frac{1}{2}, \] we get \[ T_2 = 310 \left(2\right)^{1/2}. \] \[ T_2 = 310\sqrt2. \] Using \[ \sqrt2\approx1.414, \] \[ T_2 = 310\times1.414. \] \[ T_2 = 438.3\ \text{K}. \]
Step 4: Convert into Celsius scale.
\[ T_2(^\circ\text{C}) = 438.3-273. \] \[ T_2 = 165.3^\circ\text{C}. \]
Step 5: Final conclusion.
Therefore, the final temperature of the gas is \[ \boxed{165.3^\circ\text{C}} \] Hence, the correct option is \[ \boxed{(1)} \]
Initial temperature: \[ T_1=37^\circ\text{C}=310\ \text{K}. \] Ratio of specific heats: \[ \gamma=1.5=\frac{3}{2}. \] The gas is compressed to half its original volume, therefore \[ V_2=\frac{V_1}{2}. \] Hence, \[ \frac{V_1}{V_2}=2. \]
Step 2: Use the adiabatic relation.
For an adiabatic process, \[ TV^{\gamma-1}=\text{constant}. \] Therefore, \[ T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}. \] Hence, \[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1}. \]
Step 3: Substitute the given values.
Since \[ \gamma-1 = \frac{3}{2}-1 = \frac{1}{2}, \] we get \[ T_2 = 310 \left(2\right)^{1/2}. \] \[ T_2 = 310\sqrt2. \] Using \[ \sqrt2\approx1.414, \] \[ T_2 = 310\times1.414. \] \[ T_2 = 438.3\ \text{K}. \]
Step 4: Convert into Celsius scale.
\[ T_2(^\circ\text{C}) = 438.3-273. \] \[ T_2 = 165.3^\circ\text{C}. \]
Step 5: Final conclusion.
Therefore, the final temperature of the gas is \[ \boxed{165.3^\circ\text{C}} \] Hence, the correct option is \[ \boxed{(1)} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Thermodynamics Questions
- A solid sphere at a temperature of 400 K radiates a power $P$. If the radius of the sphere is halved and its absolute temperature is doubled, then the power radiated by it is
- When a polyatomic gas is heated at constant pressure, the percentage of heat given to the gas that is converted into external work is
(Ratio of the specific heat capacities of the gas = $\frac{4}{3}$) - The thermodynamic process in which the change in internal energy of the system becomes zero is
- The efficiency of a Carnot engine operating between the temperatures 1000 K and 300 K is
- A closed vessel contains a gas at a pressure $P$. If 50% of the mass of the gas is removed and rms speed of the gas molecules is increased by 20%, then the pressure of the remaining gas is
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions