The fugacity of liquid water at 298 K is approximately 3171 P_a. Considering the ideal heat of vaporisation as 43723 J/gm mole, its fugacity at 300 K would be

ANS:B - 3567P_a

To find the fugacity of liquid water at 300 K, we can use the Clausius-Clapeyron equation: ln(P1​P2​​)=RΔHvap​​(T1​1​−T2​1​) Where:

• P1​ and P2​ are the pressures at temperatures T1​ and T2​ respectively.
• vapΔHvap​ is the molar enthalpy of vaporization.
• R is the gas constant.
• T1​ and T2​ are the temperatures.

Given:

• 1=3171P1​=3171 Pa (at 1=298T1​=298 K)
• 2=300T2​=300 K
• vap=43723ΔHvap​=43723 J/g mol

First, we convert the enthalpy of vaporization to J/mol: vap=43723 J/g molΔHvap​=43723J/g mol Then, we calculate the fugacity at 300 K: ln⁡(23171)=43723 J/mol8.314 J/(mol K)(1298 K−1300 K)ln(3171P2​​)=8.314J/(mol K)43723J/mol​(298K1​−300K1​) ln⁡(23171)=53.854 Kln(3171P2​​)=53.854K 23171=53.8543171P2​​=e53.854 2=3171×53.854P2​=3171×e53.854 2≈1.015×105 PaP2​≈1.015×105Pa So, the fugacity of liquid water at 300 K is approximately 1.015×1051.015×105 Pa. Therefore, among the options provided, the closest answer is "1.01 x 10^5 Pa".