**13.3 Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?

(b) Which is true: T₁ > T₂ or T₁ < T₂?

(c)** What is the value of PV/T where the curves meet on the y-axis?

(d) If we obtained similar plots for 1.00×10^–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen

Yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mo1^–1 K^–1.)

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Payal Gupta | Contributor-Level 10

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13.3 (a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e. $\frac{P V}{T}$ = $?$ R, (where $?$ is the number of moles and R is the universal gas constant) is a constant quantity. It is not dependent on the pressure of the gas. The curve of the gas at temperature $T_{1}$ is closer to the dotted plot than the curve of the gas at temperature $T_{2 .}$

(b) A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore $T_{1}$ > $T_{2}$ is true for the given plot.

(c) The value of the ratio $\frac{P V}{T}$ , where the two curves meet, is $?$ R. This is because the ideal gas e

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(b) A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore $T_{1}$ > $T_{2}$ is true for the given plot.

(c) The value of the ratio $\frac{P V}{T}$ , where the two curves meet, is $?$ R. This is because the ideal gas equation is given as $\frac{P V}{T}$ = $?$ R. From the given data, we have

Molecular mass of oxygen = 32 g

Mass of oxygen = 1 $\times 10^{- 3}$ = 1 g

R = 8.314 J/mole/K

Then $\frac{P V}{T}$ = $\frac{1}{32} \times 8.314$ = 0.26 J/k

Hence, the value of the ratio $\frac{P V}{T}$ , where the curves meet on the y-axis, is 0.26 J/K.

(d) If we obtain similar plots for 1.00 $\times 10^{- 3}$ kg of hydrogen, then we will not get the same values of $\frac{P V}{T}$ at the point where the curves meet on the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from the oxygen (32.0 u). We have:

$\frac{P V}{T} = 0.26 J / K$ .

Given, molecular mass (M) of $H_{2}$ = 2.02 u

$\frac{P V}{T}$ = $?$ R at constant temperature, where $?$ = $\frac{m}{M}$

M = mass of $H_{2}$

Then, m = $\frac{P V}{T} \times \frac{M}{R}$ = $\frac{0.26 \times 2.02}{8.31}$ = 6.3 $\times 10^{- 2}$ g = 6.3 $\times 10^{- 5}$ kg

Hence, 6.3 $\times 10^{- 5}$ kg of $H_{2}$ will yield the same value of $\frac{P V}{T}$

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13.3 (a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e. <math><mfrac><mrow><mrow><mi>P</mi><mi>V</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac></math> = <math><mi>?</mi></math> R, (where <math><mi>?</mi></math> is the number of moles and R is the universal gas constant) is a constant quantity. It is not dependent on the pressure of the gas. The curve of the gas at temperature <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> is closer to the dotted plot than the curve of the gas at temperature <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn><mo>.</mo></mrow></mrow></msub></math>(b) A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> > <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> is true for the given plot.(c) The value of the ratio <math><mfrac><mrow><mrow><mi>P</mi><mi>V</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac></math> , where the two curves meet, is <math><mi>?</mi></math> R. This is because the ideal gas equation is given as <math><mfrac><mrow><mrow><mi>P</mi><mi>V</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac></math> = <math><mi>?</mi></math> R. From the given data, we haveMolecular mass of oxygen = 32 gMass of oxygen = 1 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math> = 1 gR = 8.314 J/mole/KThen <math><mfrac><mrow><mrow><mi>P</mi><mi>V</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>32</mn></mrow></mrow></mfrac><mo>×</mo><mn>8.314</mn></math> = 0.26 J/kHence, the value of the ratio <math><mfrac><mrow><mrow><mi>P</mi><mi>V</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac></math> , where the curves meet on the y-axis, is 0.26 J/K.(d) If we obtain similar plots for 1.00 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math> kg of hydrogen, then we will not get the same values of <math><mfrac><mrow><mrow><mi>P</mi><mi>V</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac></math> at the point where the curves meet on the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from the oxygen (32.0 u). We have:<math><mfrac><mrow><mrow><mi>P</mi><mi>V</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac><mo>=</mo><mn>0.26</mn><mi mathvariant="normal"></mi><mi mathvariant="normal">J</mi><mo>/</mo><mi mathvariant="normal">K</mi></math> .Given, molecular mass (M) of <math><msub><mrow><mrow><mi>H</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = 2.02 u<math><mfrac><mrow><mrow><mi>P</mi><mi>V</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac></math> = <math><mi>?</mi></math> R at constant temperature, where <math><mi>?</mi></math> = <math><mfrac><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>M</mi></mrow></mrow></mfrac></math>M = mass of <math><msub><mrow><mrow><mi>H</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math>Then, m = <math><mfrac><mrow><mrow><mi>P</mi><mi>V</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac><mo>×</mo><mfrac><mrow><mrow><mi>M</mi></mrow></mrow><mrow><mrow><mi>R</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>0.26</mn><mi></mi><mo>×</mo><mn>2.02</mn></mrow></mrow><mrow><mrow><mn>8.31</mn></mrow></mrow></mfrac></math> = 6.3 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math> g = 6.3 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>5</mn></mrow></mrow></msup></math> kgHence, 6.3 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>5</mn></mrow></mrow></msup></math> kg of <math><msub><mrow><mrow><mi>H</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> will yield the same value of <math><mi></mi><mfrac><mrow><mrow><mi>P</mi><mi>V</mi></mrow></mrow><mrow><mrow><mi>T</mi></mrow></mrow></mfrac></math>

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