In given figure shows plot of $$\dfrac{PV}{T}$$ 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 $$\dfrac{PV}{T}$$ 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 $$\dfrac{PV}{T}$$ at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of $$\dfrac{PV}{T}$$(for low pressure high temperature region of the plot)? ($$Molecular;\ mass\; of\; H_2 = 2.02 u, of$$

$$O_2 = 32.0 u, R = 8.31 J mo1^{–1} K{–1})$$

Asked by Pragya Singh | 1 year ago |  148

##### Solution :-

(a) dotted plot is parallel to X-axis, signifying that nR [$$\dfrac{PV}{T}$$ = nR] is independent of P. Thus it is representing ideal gas behaviour

(b) the graph at temperature T1 is closer to ideal behaviour (because closer to dotted line) hence, T1 > T2 (higher the temperature, ideal behaviour is the higher)

(c) use PV = nRT

$$\dfrac{PV}{T}$$ = nR

Mass of the gas = 1 x 10-3 kg = 1 g

Molecular mass of O2 = 32g/mol

Hence,

Number of mole = given weight/molecular weight

$$\dfrac{1}{32}$$

So, nR = $$\dfrac{1}{32}$$ x 8.314 = 0.26 J/K

Hence,

Value of  $$\dfrac{PV}{T}$$= 0.26 J/K

(d) 1 g of H2 doesn’t represent the same number of mole

Eg. molecular mass of H2 = 2 g/mol

Hence, number of moles of H2 require is $$\dfrac{1}{32}$$ (as per the question)

Therefore,

Mass of H2 required = no. of mole of H2 x molecular mass of H2

$$\dfrac{1}{32}$$ x 2

$$\dfrac{1}{16}$$ g

= 0.0625 g

= 6.3 x 10-5 kg

Hence, 6.3 x 10-5 kg of H2 would yield the same value

Answered by Pragya Singh | 1 year ago

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