if 2 moles of real gas occupies 11.2litre volume at STP then compressibility factor of gas at STP is​

Answers 2

2 mole of gas occupies 22.4 litres of volume

Explanation:

11.2 g of gas has vapour density = 11.2

We know, Molar mass = 2 × Vapour density (VD)

Molar mass = 2 × 11.2 = 22.4 g/mol

No. of moles of gases = 22.4/ 11.2

= 2 moles

Given:

The number of moles of the gas = 2

The volume occupied by the gas at STP = 11.2 L

To Find:

The compressibility factor of gas at STP

Solution:

The compressibility factor of the gas at STP is 0.25.

The compressibility factor of a gas denoted by Z is a measure of the deviation of the gas from the behavior of an ideal gas.

Mathematically,

Z = PV/nRT

Here, P = Pressure

V = Volume occupied by the gas

n = The number of moles of the gas

R = Universal Gas Constant

T = Absolute Temperature

We know that at STP, the pressure is 1 atm and the temperature is 273.15 K.

Substituting the values,

Z = 1 X 11.2 / 2 X 0.082 X 273.15

= 11.2 / 44.80

= 0.25

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3. The following data gives the information on the observed life times in hours) of 150 components electrical components Life Time (in hours) 0-20 20 - 40 40 - 60 60 - 80 80 - 100 Frequency 15 10 35 50 40 Find the mode of distribution.​

Answer:

Mode= 72

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answer img

Answer:

mode of the given grouped data  = 72

Step-by-step explanation:

class   frequency

0-20          15

20-40        10

40-60        35

60-80        50

80-100       40

maximum frequency is 50 and the corresponding class is 60-80

so modal class is 60-80

for a grouped data, mode = [tex]l+\frac{(f1-fo)h}{2f1-fo-f2}[/tex]

where:

l is the lower limit of the modal class = 60

fo is the frequency anteceding the modal class = 35

f1 is the frequency of the modal class = 50

f2 is the frequency following the modal class = 40

and, h is the size of class interval = 20

then: mode = 60 + [tex]\frac{(50-35)20}{2(50) - 35-40}[/tex]

⇒ mode = 60 + [tex]\frac{(15)(20)}{25}[/tex]

      = 60 + 12

      = 72

hence, mode of the given grouped data  = 72

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