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InterviewSolution
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Caculate molecular K.E. of 1 g of helium at NTP. What will be its energy at `100^(@)C`? |
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Answer» Correct Answer - `8.51 xx 10^(2)J`, 11.63 xx 10^(2)J` At NTP , `P=1.013 xx 10^(5) N//m^(2), T_(0) = 273 K` `V = 22.4 litre = 22.4 xx 10^(-3) m^(3), M=4` As helium is monoatomic gas, therefore, K.E.//mole = `3/2 RT_(0) = 3/2 PV` `K.E.//gram =3/2 (PV)/(M)` `= 3/2 xx (1.013 xx 10^(5) xx 22.4 xx 10^(-3))/(4)` `E_(0) = 8.51 xx 10^(2)J` |
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