From the relation PV = RT, calculate the value of the constant R for one gram mole of an ideal gas (in `cal//K`)

Answer

2

Explanation

From `PV = RT`
`R = (pV)/(t)`
`Now, P=1atm =10^(5)N//m^(2)`
`V=22.4 litre = 22.4 xx 10^(-3)m^(3)`
`T = 273 K`
`:. R=(10^(5)xx22.4xx10^(-3))/(273) J//K`
`=(8.2)/(4.18) cal//K = 1.96 cal//K ~= 2 cal//K`.