Home
Class 11
PHYSICS
A mono atomic gas initially at 27^(@)C i...

A mono atomic gas initially at `27^(@)C` is compressed adiabatically to one eighth of its original volume. The temperature after compression will be

A

`10^(@)C`

B

`887^(@)C`

C

`927^(@)C`

D

`144^(@)C`

Text Solution

AI Generated Solution

The correct Answer is:
To find the final temperature of a monoatomic gas after it has been compressed adiabatically to one-eighth of its original volume, we can use the adiabatic relation for an ideal gas. Here’s a step-by-step solution: ### Step 1: Convert the initial temperature to Kelvin The initial temperature \( T_1 \) is given as \( 27^\circ C \). To convert this to Kelvin: \[ T_1 = 27 + 273 = 300 \, K \] ### Step 2: Identify the initial and final volumes Let the initial volume be \( V_1 \). According to the problem, the final volume \( V_2 \) after compression is: \[ V_2 = \frac{V_1}{8} \] ### Step 3: Use the adiabatic relation For an adiabatic process involving an ideal gas, the following relation holds: \[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \] where \( \gamma \) (gamma) is the heat capacity ratio. For a monoatomic gas, \( \gamma = \frac{5}{3} \). ### Step 4: Substitute the known values into the equation Substituting the known values into the adiabatic relation: \[ 300 \cdot V_1^{\frac{5}{3} - 1} = T_2 \cdot \left(\frac{V_1}{8}\right)^{\frac{5}{3} - 1} \] This simplifies to: \[ 300 \cdot V_1^{\frac{2}{3}} = T_2 \cdot \left(\frac{V_1}{8}\right)^{\frac{2}{3}} \] ### Step 5: Simplify the equation Rearranging the equation gives: \[ T_2 = 300 \cdot \frac{V_1^{\frac{2}{3}}}{\left(\frac{V_1}{8}\right)^{\frac{2}{3}}} \] This can be further simplified: \[ T_2 = 300 \cdot \frac{V_1^{\frac{2}{3}}}{\frac{V_1^{\frac{2}{3}}}{8^{\frac{2}{3}}}} = 300 \cdot 8^{\frac{2}{3}} \] ### Step 6: Calculate \( 8^{\frac{2}{3}} \) Calculating \( 8^{\frac{2}{3}} \): \[ 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4 \] ### Step 7: Final calculation of \( T_2 \) Now substituting back: \[ T_2 = 300 \cdot 4 = 1200 \, K \] ### Conclusion The final temperature \( T_2 \) after the adiabatic compression is: \[ \boxed{1200 \, K} \]

To find the final temperature of a monoatomic gas after it has been compressed adiabatically to one-eighth of its original volume, we can use the adiabatic relation for an ideal gas. Here’s a step-by-step solution: ### Step 1: Convert the initial temperature to Kelvin The initial temperature \( T_1 \) is given as \( 27^\circ C \). To convert this to Kelvin: \[ T_1 = 27 + 273 = 300 \, K \] ...
Promotional Banner

Topper's Solved these Questions

  • THERMAL PROPERTIES OF MATTER

    NARAYNA|Exercise LEVEL - II (H.W.)|19 Videos
  • TRANSMISSION OF HEAT

    NARAYNA|Exercise LEVEL-II(C.W)|27 Videos

Similar Questions

Explore conceptually related problems

A diatomic gas initially at 18^(@) is compressed adiabatically to one- eighth of its original volume. The temperature after compression will b

A diatomic gas initally at 18^(@)C is compressed adiabtically to one- eighth of its original volume. The temperature after compression will be

A monatomic gas initially at 17^@ C has suddenly compressed adiabatically to one-eighth of its original volume. The temperature after compression is

If a gas is compressed adiabatically

An ideal gas at 27^(@)C is compressed adiabatically to 8//27 of its original volume. If gamma = 5//3 , then the rise in temperature is

NARAYNA-THERMODYNAMICS-Exercise
  1. The pressure and density of a monoatmic gas (gamma=5//3) change adiaba...

    Text Solution

    |

  2. Air is filled in a motor tube at 27^(@)C and at a pressure of 8 atmosp...

    Text Solution

    |

  3. A mono atomic gas initially at 27^(@)C is compressed adiabatically to ...

    Text Solution

    |

  4. One gm mol of a diatomic gas (gamma=1.4) is compressed adiabatically s...

    Text Solution

    |

  5. A container of volume 2m^(3) is divided into two equal compartments, o...

    Text Solution

    |

  6. At 27^(@)C and pressure of 76 cm of Hg the volume of a diatomic gas is...

    Text Solution

    |

  7. The work done on a gas when it is compressed isothermally at 27^(@)C t...

    Text Solution

    |

  8. A Carnot engine has the same efficiency between 800 K to 500 K and x K...

    Text Solution

    |

  9. A Carnot's engine woeking between 27^(@)C and 127^(@)C takes up 800 J ...

    Text Solution

    |

  10. A copper block of mass 5kg slides down along a rough inclined plane of...

    Text Solution

    |

  11. A brass sphere of mass 0.2kg falls freely from a height of 20m and bou...

    Text Solution

    |

  12. A lead bullet of 10g travelling at 300m//s strikes against a block of ...

    Text Solution

    |

  13. Water falls from a height 500m, what is the rise in temperature of wat...

    Text Solution

    |

  14. A ball is dropped on a floor from a height of 2.0m. After the collisio...

    Text Solution

    |

  15. In a thermodynamic process, pressure of a fixed mass of a gas is chang...

    Text Solution

    |

  16. The density of a substance is 400kgm^(-3) and that of another substanc...

    Text Solution

    |

  17. If for hydrogen C(P) - C(V) = m and for nitrogen C(P) - C(V) = n, wher...

    Text Solution

    |

  18. 294 joules of heat is requied to rise the temperature of 2 mole of an ...

    Text Solution

    |

  19. 1672cal of heat is given to one mole of oxygen at 0^(@)C keeping the v...

    Text Solution

    |

  20. 0.5 mole of diatomic gas at 27^(@)C is heated at constant pressure so ...

    Text Solution

    |