MCQEasyJEE 2024Equation of State of Ideal Gas

JEE Physics 2024 Question with Solution

The temperature of a gas having 2.0×10252.0\times10^{25} molecules per cubic meter at 1.38atm1.38\,\text{atm} (Given, k=1.38×1023J K1k = 1.38\times10^{-23}\,\text{J K}^{-1}) is:

  • A

    500K500\,\text{K}

  • B

    200K200\,\text{K}

  • C

    100K100\,\text{K}

  • D

    300K300\,\text{K}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: number density NV=2.0×1025m3\frac{N}{V} = 2.0\times10^{25}\,\text{m}^{-3}, pressure P=1.38atmP = 1.38\,\text{atm}, and Boltzmann constant k=1.38×1023J K1k = 1.38\times10^{-23}\,\text{J K}^{-1}.

Find: the temperature TT of the gas.

Use the ideal gas law in molecular form:

PV=NkTPV = NkT

Since the number of molecules per unit volume is given, write

P=NVkTP = \frac{N}{V}kT

So,

T=P(NV)kT = \frac{P}{\left(\frac{N}{V}\right)k}

Convert pressure into pascal:

P=1.38×101325PaP = 1.38 \times 101325\,\text{Pa}

Substitute the values:

T=1.38×1013252.0×1025×1.38×1023T = \frac{1.38 \times 101325}{2.0\times10^{25} \times 1.38\times10^{-23}}

Now simplify:

T=1.39657×1052.76×102T = \frac{1.39657\times10^{5}}{2.76\times10^{2}} T500KT \approx 500\,\text{K}

Therefore, the temperature of the gas is 500K500\,\text{K}. The correct option is A.

Common mistakes

  • Using PV=nRTPV = nRT directly with number of molecules is incorrect because nn is in moles, not molecules. Use PV=NkTPV = NkT or convert molecules into moles first.

  • Treating 2.0×10252.0\times10^{25} as the total number of molecules instead of number density is wrong. The given quantity is molecules per cubic meter, so use NV\frac{N}{V} in the formula.

  • Not converting pressure from atmosphere to pascal leads to a wrong numerical value. Replace 1.38atm1.38\,\text{atm} by 1.38×101325Pa1.38\times101325\,\text{Pa} before substitution.

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