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Lana71 [14]
4 years ago
5

A sample of gas at 25ºc has a volume of 11 l and exerts a pressure of 660 mm hg. how many moles of gas are in the sample?

Chemistry
1 answer:
san4es73 [151]4 years ago
5 0
  The  moles of  the gas    in the sample  is 0.391  moles

    calculation
by use  of  ideal  gas  equation, that  is Pv=nRT
where  n  is number  of moles
           P(pressure)= 660  mmhg
           R(gas  constant)  =  62.364 l.mmhg/mol.K
           T(temperature)= 25 +273 =   298 k

by  making  n  the subject of the  formula

n= Pv/ RT

n is therefore=   (660mm hg  x11 L)/(  62.364 L.mmhg/mol.k  x298 K)  = 0.391  moles
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Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]

Where,

\Delta H_v is the Heat of vaoprization (J/mol)

T_b is the normal boiling point of the gas (K)

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For diethyl ether:

T_b=307.4\ K

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Applying the above equation to find heat of vaporization as:

\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]

\Delta H_v=26400 J/mol

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

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