Finding the original pressure, thermodynamics

[jweie29nh]

Homework Statement

Freon 12 is contained in a sealed glass container at 50°C. As it is cooled, vapor droplets arc noted
condensing on the sidewalls at 20°C. Find the original pressure in the container.

Homework Equations

The Attempt at a Solution

Looking at the Freon 12 table I subtracted the given pressures from 50 degrees and 20 degrees to get 652 kpa. I'm pretty sure the this is an appropriate solution. More likely there is another approach to solving this problem.

 

[Charles Link]

At $T=50^{\circ}$ C , it is all vaporized so the pressure is less than the vapor pressure at that temperature. I'm presuming at $T=21^{\circ}$ C there still were no droplets forming. Would not the approximate answer be the vapor pressure at $T=20^{\circ}$ C which is $P=566$ kPa, or just slightly more than this? $\\$ As Homework Helpers we are not supposed to supply the answer, but this one involves an interpretation, and I think it has a very simple answer, but the OP still needs to determine if what I have proposed is indeed correct. $\\$ See also: https://edisciplinas.usp.br/pluginfile.php/283793/mod_resource/content/2/R12 Properties.pdf $\\$ And meanwhile, I think the solution that you have proposed is incorrect. The first temperature they chose to present could have been any temperature above $T=20^{\circ}$ C , including $T=21^{\circ}$ C . Taking $P(21^{\circ} \, C)-P(20^{\circ} \, C)$ would be incorrect.

 

[Chestermiller]

If the vapor pressure of an ideal gas is 566 kPa at 20 C, what is the pressure of the gas at 50 C (assuming the volume is constant)?

 

[Charles Link]

If the vapor pressure of an ideal gas is 566 kPa at 20 C, what is the pressure of the gas at 50 C (assuming the volume is constant)?

Thank you @Chestermiller . I overlooked this second part of the problem. Hopefully the OP returns to also figure out how to work it.