We have 0.5 g of hydrogen gas in a cubic chamber of size 3cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state? (Hydrogen molecules can be consider as spheres of radius 1 A^o ).

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Payal Gupta | Contributor-Level 10

4 months ago

This is a short answer type question as classified in NCERT Exemplar

Radius = 1A^o

Volume of hydrogen molecules = 4/3 $π$ r³

= 4/3 (3.14) (10^-10)³ $\approx 4 \times 10^{- 30}$ m³

Number of moles of H₂ = mass/molecular mass=0.5/2=0.25

Molecules of H₂ present = number of moles of H₂present $\times 6.023 \times 10^{23}$

= 0.25 $\times 6.023 \times 10^{23}$

So volume of molecules present = molecule number $\times$ volume of each molecules

= 0.25 $\times 6.0233 \times 10^{23} \times 4 \times 10^{23}$

$\approx$ 6 $\times 10^{- 7} m$ ³

P_iV_i= P_fV_f

V_{f
$(\frac{P_{i}}{p_{f}}) V$}= _i= $\frac{1}{100} (3 \times 10^{- 2})$ ³

V_f= 2.7 $\times 10^{- 7} m$ ³

This is a short answer type question as classified in NCERT ExemplarRadius = 1AoVolume of hydrogen molecules = 4/3 <math> <mi>π</mi> </math> r3= 4/3 (3.14) (10-10)3 <math> <mo>≈</mo> <mn>4</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>30</mn> </mrow> </mrow> </msup> </math> m3Number of moles of H2 = mass/molecular mass=0.5/2=0.25Molecules of H2 present = number of moles of H2 present <math> <mo>×</mo> <mn>6.023</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> </math> = 0.25 <math> <mo>×</mo> <mn>6.023</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> </math> So volume of molecules present = molecule number <math> <mo>×</mo> </math> volume of each molecules= 0.25 <math> <mo>×</mo> <mn>6.0233</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>4</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> </math> <math> <mo>≈</mo> </math> 6 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>7</mn> </mrow> </mrow> </msup> <mi>m</mi> </math> 3PiVi= PfVfVf <math> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> <mrow> <mrow> <mi>f</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> <mi>V</mi> </math> = i= <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>100</mn> </mrow> </mrow> </mfrac> <mo> (</mo> <mn>3</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> <mo>)</mo> </math> 3Vf= 2.7 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>7</mn> </mrow> </mrow> </msup> <mi>m</mi> </math> 3

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