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CBSE Class 11-science Answered

6th?
Asked by smanishkumar2002 | 02 Aug, 2018, 09:23: PM
answered-by-expert Expert Answer
Question: The critical temperature and pressure of the CO2 gas are 304.2 K and 72.9 atm respectively. Calculate the radius of the CO2 molecule assuming it to behave van der Wall’s gas.
 
Solution:
 
Given:
 
Tc = 304.2 K
 
Pc = 72.9 atm
 
We have,
 
straight T subscript straight c equals space fraction numerator 8 straight a over denominator 27 Rb end fraction straight P subscript straight c space end subscript equals fraction numerator straight a over denominator 27 straight b squared end fraction straight T subscript straight c over straight P subscript straight c equals fraction numerator fraction numerator 8 straight a over denominator 27 Rb end fraction over denominator fraction numerator straight a over denominator 27 straight b squared end fraction end fraction space space space space space equals fraction numerator 8 up diagonal strike straight a over denominator up diagonal strike 27 straight R up diagonal strike straight b end fraction cross times fraction numerator up diagonal strike 27 straight b to the power of up diagonal strike 2 end exponent over denominator up diagonal strike straight a end fraction straight T subscript straight c over straight P subscript straight c equals fraction numerator 8 straight b over denominator straight R end fraction OR straight b space equals straight R over 8 straight T subscript straight c over straight P subscript straight c space space space equals fraction numerator 0.082 cross times 304.2 over denominator 8 cross times 72.9 end fraction space straight b equals 0.04277 space lit divided by mol space space space space equals 42.77 space cm cubed straight b space equals 4 straight N subscript straight A 4 over 3 πr cubed therefore straight r cubed equals space space fraction numerator 3 cross times 42.77 cross times 10 to the power of negative 23 end exponent over denominator 16 cross times 6.023 cross times 3.14 end fraction space space space straight r cubed equals 0.424 cross times 10 to the power of negative 23 end exponent straight r space equals open parentheses 4.24 close parentheses cross times 10 to the power of negative 24 end exponent straight r equals open parentheses 4.24 close parentheses to the power of bevelled 1 third end exponent cross times 10 to the power of negative 8 end exponent space space cm straight r equals space 1.62 cross times 10 to the power of negative 8 end exponent space space cm straight r equals 1.62 space to the power of ring operator straight A
The radius of CO2 is 1.62 °A.
Answered by Varsha | 03 Aug, 2018, 01:16: PM

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