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Heat and Thermodynamics question

2022 · 29 Jun · Shift 2 · Q54
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  5. /2022 · 29 Jun · Shift 2 · Q54

Heat and Thermodynamics question

2022 · 29 Jun · Shift 2 · Q54

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A vessel contains 16g of hydrogen and 128g of oxygen at standard temperature and pressure. The volume of the vessel in cm3 is :
  1. A
    72 ×\times× 105
  2. B
    32 ×\times× 105
  3. C
    27 ×\times× 104
  4. D
    54 ×\times× 104
View written solutionFree

Correct answer: C

  1. Find the number of moles of each gas

    • Hydrogen: molecular mass of H2=2 g/molH_2 = 2\,\text{g/mol}H2​=2g/mol nH2=162=8 moln_{H_2} = \frac{16}{2} = 8\,\text{mol}nH2​​=216​=8mol

    • Oxygen: molecular mass of O2=32 g/molO_2 = 32\,\text{g/mol}O2​=32g/mol nO2=12832=4 moln_{O_2} = \frac{128}{32} = 4\,\text{mol}nO2​​=32128​=4mol

  2. Total number of moles in the vessel

    Since both gases are simply present together, ntotal=8+4=12 moln_{\text{total}} = 8 + 4 = 12\,\text{mol}ntotal​=8+4=12mol

  3. Use molar volume at STP

    At standard temperature and pressure, 111 mole of an ideal gas occupies 22.4 L=22.4×103 cm322.4\,\text{L} = 22.4 \times 10^3\,\text{cm}^322.4L=22.4×103cm3

    Therefore, volume of 121212 moles is V=12×22.4 L=268.8 LV = 12 \times 22.4\,\text{L} = 268.8\,\text{L}V=12×22.4L=268.8L

  4. Convert into cm3\text{cm}^3cm3

    268.8 L=268.8×103 cm3=2.688×105 cm3268.8\,\text{L} = 268.8 \times 10^3\,\text{cm}^3 = 2.688 \times 10^5\,\text{cm}^3268.8L=268.8×103cm3=2.688×105cm3

    This can be written as 2.688×105=26.88×104≈27×104 cm32.688 \times 10^5 = 26.88 \times 10^4 \approx 27 \times 10^4\,\text{cm}^32.688×105=26.88×104≈27×104cm3

  5. Match with the options

    The closest and correct option is: 27×104 cm3\boxed{27 \times 10^4\,\text{cm}^3}27×104cm3​

    So, Option C is correct.

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