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Gaseous State question

2023 · Shift 1 · Q9
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Gaseous State question

2023 · Shift 1 · Q9

JEE AdvancedChemistryGaseous StateNumerical+4 / −1
A gas has a compressibility factor of 0.5 and a molar volume of 0.4 dm3 mol−10.4 ~\mathrm{dm}^3 \mathrm{~mol}^{-1}0.4 dm3 mol−1 at a temperature of 800 K800 \mathrm{~K}800 K and pressure x\mathbf{x}x atm. If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be y dm3 mol−1\mathbf{y} ~\mathrm{dm}^3 \mathrm{~mol}^{-1}y dm3 mol−1. The value of x/y\mathbf{x} / \mathbf{y}x/y is ‾\underline{\hspace{2cm}}​. [Use: Gas constant, R=8×10−2 L\mathrm{R}=8 \times 10^{-2} \mathrm{~L}R=8×10−2 L atm K−1 mol−1\mathrm{K}^{-1} \mathrm{~mol}^{-1}K−1 mol−1 ]
Numerical answer
View written solutionFree

Correct answer: 100

  1. Use the definition of compressibility factor

    Z=PVmRTZ=\frac{PV_m}{RT}Z=RTPVm​​

    Given:

    • Z=0.5Z=0.5Z=0.5
    • Vm=0.4 dm3 mol−1=0.4 L mol−1V_m=0.4\,\mathrm{dm^3\,mol^{-1}}=0.4\,\mathrm{L\,mol^{-1}}Vm​=0.4dm3mol−1=0.4Lmol−1
    • T=800 KT=800\,\mathrm{K}T=800K
    • R=8×10−2 L atm K−1 mol−1R=8\times 10^{-2}\,\mathrm{L\,atm\,K^{-1}\,mol^{-1}}R=8×10−2LatmK−1mol−1

  2. Find the pressure xxx

    0.5=x×0.4(8×10−2)×8000.5=\frac{x\times 0.4}{(8\times 10^{-2})\times 800}0.5=(8×10−2)×800x×0.4​

    First calculate RTRTRT:

    (8×10−2)×800=64(8\times 10^{-2})\times 800=64(8×10−2)×800=64

    So,

    0.5=0.4x640.5=\frac{0.4x}{64}0.5=640.4x​

    0.5×64=0.4x0.5\times 64=0.4x0.5×64=0.4x

    32=0.4x32=0.4x32=0.4x

    x=320.4=80 atmx=\frac{32}{0.4}=80\,\text{atm}x=0.432​=80atm

  3. Find the ideal gas molar volume yyy at same TTT and PPP

    For an ideal gas, Z=1Z=1Z=1, so

    PVm=RTPV_m=RTPVm​=RT

    Hence,

    y=RTxy=\frac{RT}{x}y=xRT​

    y=6480=0.8 dm3 mol−1y=\frac{64}{80}=0.8\,\mathrm{dm^3\,mol^{-1}}y=8064​=0.8dm3mol−1

  4. Calculate x/yx/yx/y

    xy=800.8=100\frac{x}{y}=\frac{80}{0.8}=100yx​=0.880​=100

  5. Final Answer

    100\boxed{100}100​

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