Class 11 Chemistry Gases & Solutions States of Matter

What is the value of the absolute zero of temperature on their temperature scale?

(a)

$233.33_{o}$C

(b)

$−233.33_{o}$C

(c)

$−233.33_{o}$N

(d)

$233.33_{o}$N

Correct answer: (c)

Solution: Given,

Here $N$ denotes Newton Degree. One Newton Degree is equal to $10033 _{o}C$ .

Melting point$=0_{o}$N

Boiling point$=100_{o}$N

given the value A $PV_{m}=28$ at $0_{o}$N

$PV_{m}=40$ at $100_{o}$N

Using ideal gas equation for reduced volume, $n=1$

also taking the absolute value as x.

$∴28=R(x+0)$ (at $0_{o}$N)

$⇒Rx=28$ …….$(1)$

at $100_{o}$N

$40=R(x+100)$

$⇒40=Rx+R×100$

$R×100=40−28$

$R=10012 =0.12$

On putting back in equation $(1)$

we get $x=1228×100 $

$⇒x=233.33$

In general absoulte zero is $0K$ or $−273_{o}C$

$[K]=[_{o}C]+273$

Similarly, here

$[K]=[_{o}N]+233$

$∴$ Absolute Zero is $−x$

$−x=−233.33$

Option C.

Here $N$ denotes Newton Degree. One Newton Degree is equal to $10033 _{o}C$ .

Melting point$=0_{o}$N

Boiling point$=100_{o}$N

given the value A $PV_{m}=28$ at $0_{o}$N

$PV_{m}=40$ at $100_{o}$N

Using ideal gas equation for reduced volume, $n=1$

also taking the absolute value as x.

$∴28=R(x+0)$ (at $0_{o}$N)

$⇒Rx=28$ …….$(1)$

at $100_{o}$N

$40=R(x+100)$

$⇒40=Rx+R×100$

$R×100=40−28$

$R=10012 =0.12$

On putting back in equation $(1)$

we get $x=1228×100 $

$⇒x=233.33$

In general absoulte zero is $0K$ or $−273_{o}C$

$[K]=[_{o}C]+273$

Similarly, here

$[K]=[_{o}N]+233$

$∴$ Absolute Zero is $−x$

$−x=−233.33$

Option C.

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