In given figure shows plot of \( \dfrac{PV}{T}\) versus P for \( 1.00×10^{-3}\) kg of oxygen gas at two different temperatures.

Figure shows plot of PV/T versus P for 1.00 × 10^-3 kg of oxygen gas at two  different temperatures. - Sarthaks eConnect | Largest Online Education  Community

(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

1 Answer

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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