At `27^@C` and at a pressure of 76 cmHg `100cm^2` of a gas is collected over water surface. The space occupied by the gas is saturated with water vapour. Maximum vapour pressure of water at `27^@C` is 17.4mmHg. What will be the volume of dry gas at STP?

Explanation

Let the volume of the dry gas at STP `=V_2 cm ^3`, pressure `p_2=75 cmHg` and temperature `T_2=0^@C=273K`. Given `V_1=100 cm^3,p_1=76-1.74=74.26 cmHg` and `T_1=27+273=300 K`.
Hence from the equation of state
`(p_1V_1)/T_1=(p_2V_2)/T_2or,(100times74.26)/300=(V_2times76)/273`
or,`V_2=(273times74.26)/(3times76)=88.92cm^3`
`therefore` At STP the volume of dry gas will be `88.92 cm^3`.