CBSE Class 11-science Answered

14th?

Asked by smanishkumar2002 | 04 Aug, 2018, 05:41: AM

Expert Answer

Here we have two cases

Case I

Given :

P= 1 atm

w= 24 g

T = (t+273)K

V = V litre

Using the gas equation,

PV space equals space straight w over straight m space RT 1 space cross times space straight V space equals space 24 over straight m space straight R space left parenthesis straight T space plus space 273 right parenthesis.............................. left parenthesis 1 right parenthesis

Case II:

Given:

T = (T + 283)K

P =

1 plus fraction numerator 20 over denominator 100 space end fraction equals space 1.2 space atm

w = 24 g

V= V Litre

Using the gas equation,

PV space equals space straight w over straight m space RT 1.2 space cross times space straight V space equals space 24 over straight m space straight R space left parenthesis straight T space plus space 283 right parenthesis.............................. left parenthesis 2 right parenthesis

By equating equation (1) and (2)

fraction numerator 1.2 over denominator 1 end fraction space equals fraction numerator left parenthesis T space plus 283 right parenthesis space over denominator left parenthesis T space plus 273 right parenthesis end fraction therefore space 1.2 space T space plus space 327.6 space equals space T space plus space 283 0.2 T space equals space minus 44.6 T space equals space minus 223 degree C space equals space 50.15 K A l s o space F r o m space left parenthesis 1 right parenthesis comma space o n space s u b s t i t u t i n g space t space a n d space m equals 120 space w e space g e t space 1 space cross times space V space equals space fraction numerator 24 over denominator 120 space end fraction cross times 0.082 cross times 50.15 V space equals space 0.82 space l i t r e

Answered by Ramandeep | 06 Aug, 2018, 11:57: AM

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