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Heat and Thermodynamics question

2025 · 29 Jan · Shift 1 · Q67
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Heat and Thermodynamics question

2025 · 29 Jan · Shift 1 · Q67

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
A container of fixed volume contains a gas at 27°C. To double the pressure of the gas, the temperature of gas should be raised to ‾\underline{\hspace{2cm}}​ °C.
Numerical answer
View written solutionFree

Correct answer: 327

  1. For a gas in a fixed volume container, pressure is directly proportional to absolute temperature:

P∝TP \propto TP∝T

So,

P2P1=T2T1\frac{P_2}{P_1} = \frac{T_2}{T_1}P1​P2​​=T1​T2​​

  1. Initial temperature is given as 27∘C27^\circ\text{C}27∘C. Convert it to Kelvin:

T1=27+273=300 KT_1 = 27 + 273 = 300\,\text{K}T1​=27+273=300K

  1. The pressure is to be doubled:

P2=2P1P_2 = 2P_1P2​=2P1​

Therefore,

P2P1=2=T2T1\frac{P_2}{P_1} = 2 = \frac{T_2}{T_1}P1​P2​​=2=T1​T2​​

So,

T2=2T1=2×300=600 KT_2 = 2T_1 = 2 \times 300 = 600\,\text{K}T2​=2T1​=2×300=600K

  1. Convert back to Celsius:

t2=600−273=327∘Ct_2 = 600 - 273 = 327^\circ\text{C}t2​=600−273=327∘C

  1. Hence, the temperature should be raised to:

327∘C\boxed{327^\circ\text{C}}327∘C​

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