In the cylinder under the piston there is water vapor at a temperature of 100 ° C

In the cylinder under the piston there is water vapor at a temperature of 100 ° C and a pressure of 40 kPa. What will be the vapor pressure in the cylinder if its volume is isothermally reduced by 5 times?

The state of the gas is described by the Mendeleev-Clapeyron equation:

p * v = m / M * R * T, where p is pressure, V is volume, m is gas mass, M is molar mass, R is universal gas constant, T is temperature.

We write this equation for the first and second states:

p1 * v1 = m1 / M * R * T1;

p2 * v2 = m2 / M * R * T2.

We divide the first equation by the second (m1 = m2; T1 = T2 – according to the problem statement):

p1 * v1 = p2 * v2;

p2 = p1 * v1 / v2.

Since the volume has decreased 5 times:

v1 / v2 = 5.

p2 = 5 * p1 = 5 * 40 = 200 kPa.