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12.Kinetic Theory of Gases
normal
The energy of a gas/litre is $300$ joules, then its pressure will be
A
$3 \times {10^5}\,N/{m^2}$
B
$6 \times {10^5}\,N/{m^2}$
C
${10^5}\,N/{m^2}$
D
$2 \times {10^5}\,N/{m^2}$
Solution
(d) Energy $ = 300\,J/litre = 300 \times {10^3}\,J/{m^3}$
$P = \frac{2}{3}E = \frac{{2 \times 300 \times {{10}^3}}}{3} = 2 \times {10^5}\,N/{m^2}$
Standard 11
Physics