A)
B)
C)
D)
Correct Answer: C
Solution :
[c] Mean KE of gas molecule \[E=\frac{3}{2}kT=\frac{3}{2}k(t+273)\] |
where \[T=\] temperature is in kelvin and \[t=\] is in centigrade. |
\[\therefore \] \[\,E=\frac{3}{2}kt+\frac{3}{2}\,\times 273\,k\] |
[k = Boltzmann's constant] |
By comparing this equation with standard equation of straight line |
\[y=mx+c,\] we get \[m=\frac{3}{2}k\] and \[c=\frac{3}{2}\,273\,k\]. |
So, the graph between E and t will be straight line with positive intercept on y-axis and positive slope with r-axis. |
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