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Heat and Thermodynamics question

2025 · 3 Apr · Shift 1 · Q51
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Heat and Thermodynamics question

2025 · 3 Apr · Shift 1 · Q51

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800 cm3800 \mathrm{~cm}^3800 cm3 and temperature 27∘C27^{\circ} \mathrm{C}27∘C. The change in temperature when the gas is adiabatically compressed to 200 cm3200 \mathrm{~cm}^3200 cm3 is: (Take γ=1.5;γ\gamma=1.5 ; \gammaγ=1.5;γ is the ratio of specific heats at constant pressure and at constant volume)
  1. A
    300 K
  2. B
    600 K
  3. C
    327 K
  4. D
    522 K
View written solutionFree

Correct answer: A

  1. Identify the process

The walls are thermally non-conducting, so the compression is adiabatic.

For an adiabatic process of an ideal gas:

TVγ−1=constantTV^{\gamma-1}=\text{constant}TVγ−1=constant

  1. Write the given data
  • Initial volume: V1=800 cm3V_1=800\,\text{cm}^3V1​=800cm3
  • Final volume: V2=200 cm3V_2=200\,\text{cm}^3V2​=200cm3
  • Initial temperature: T1=27∘C=300 KT_1=27^\circ C = 300\,KT1​=27∘C=300K
  • Ratio of specific heats: γ=1.5\gamma=1.5γ=1.5
  1. Apply the adiabatic relation

T1V1γ−1=T2V2γ−1T_1 V_1^{\gamma-1}=T_2 V_2^{\gamma-1}T1​V1γ−1​=T2​V2γ−1​

So,

T2=T1(V1V2)γ−1T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}T2​=T1​(V2​V1​​)γ−1

Substitute the values:

T2=300(800200)1.5−1T_2=300\left(\frac{800}{200}\right)^{1.5-1}T2​=300(200800​)1.5−1

T2=300(4)0.5T_2=300(4)^{0.5}T2​=300(4)0.5

T2=300×2=600 KT_2=300\times 2=600\,KT2​=300×2=600K

  1. Find the change in temperature

ΔT=T2−T1=600−300=300 K\Delta T = T_2-T_1=600-300=300\,KΔT=T2​−T1​=600−300=300K

  1. Match with the options

The change in temperature is:

300 K\boxed{300\,K}300K​

So the correct option is A.

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