## Chemistry 10th Edition

Given, n= 2.54 mol T= (273+45) K = 318 K V= 12.75 L Using Ideal gas equation, P = $\frac{nRT}{V}$ = $\frac{2.54mol(0.0821\frac{L.atm}{mol.K})318K}{12.75L}$ = 5.20 atm