There is humid air in a vessel at a temperature of 20 degrees and a pressure of 760 mm

There is humid air in a vessel at a temperature of 20 degrees and a pressure of 760 mm. What is the pressure of dry air in the vessel. What will be the air pressure if the temperature is not changed, and the volume is reduced by 2 times

Given:
T1 = 20 C = 20 + 273 = 293 K
p1 = 760 mm Hg. column – this is the norm atmospheric pressure, therefore p1 = 10 ^ 5 Pa
V1
T1 = T2
V2 = V1 / 2
For gas, the ratio p1 * V1 / T1 = p2 * V2 / T2
substitute in the equation:
10 ^ 5 * V1 / 293 = p2 * V1 / 2 * 293
p2 = 2 * 10 ^ 5 Pa