For the solution of the gases w, x, y and z in water at 298 K, the Henrys law constants (KH) are 0.5, 2, 35 and 40 kbar, respectively. The correct plot for the given data is :

Question

For the solution of the gases w, x, y and z in water at 298 K, the Henrys law constants (KH) are 0.5, 2, 35 and 40 kbar, respectively. The correct plot for the given data is : (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

From Henry’s law Pgas = KH.Xgas Pgas = KH(1 -XH2O ) Pgas = KH - KH XH2O Comparing this equation with straight line equation y = mx + c

Q. For the solution of the gases $w, x, y$ and $z$ in water at $298 K$ , the Henrys law constants $(K_{H})$ are $0.5, 2, 35$ and $40 kba r$ , respectively. The correct plot for the given data is :

From Henry’s law
P_gas = K_H.X_gas
P_gas = K_H(1 -X_H₂_O )
P_gas = K_H - K_H X_H₂_O
Comparing this equation with straight line equation y = mx + c

Q. For the solution of the gases w,x,y and z in water at 298K, the Henrys law constants (KH​) are 0.5,2,35 and 40kbar, respectively. The correct plot for the given data is :

From Henry’s law Pgas = KH.Xgas Pgas = KH(1 -XH2O ) Pgas = KH - KH XH2O Comparing this equation with straight line equation y = mx + c

Q. For the solution of the gases $w, x, y$ and $z$ in water at $298 K$ , the Henrys law constants $(K_{H})$ are $0.5, 2, 35$ and $40 kba r$ , respectively. The correct plot for the given data is :

From Henry’s law
P_gas = K_H.X_gas
P_gas = K_H(1 -X_H₂_O )
P_gas = K_H - K_H X_H₂_O
Comparing this equation with straight line equation y = mx + c