Question

using perfect gas equation determine the value of R.given that 1g molecule of a gas at S. T. P occupies 22.4 l
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Answer 1

Answer:

Here, `R = ? V = 22.4` litre `= 22.4 xx 10^(3) m^(3)`

at S.T.P, `T =273 K`, and

`P=1` atmosphere `= 1.013 xx 10^(5)N//m^(2)`

For one mole of gas,

`PV =RT`

`R = (PV)/(T) = (1.013 xx 10^(5) xx 22.4 xx 10^(-3))/(273)`

`= 8.31 J mole^(-1) K^(-1)`.

Explanation: