Given a gas under #600*"mm Hg"# pressure, and at #303.15*K# temperature, that occupies a volume of #2.02*L#, what volume will it occupy at #273.15*K# temperature, under #750*"mm Hg"# pressure?

Question

Henry Antrim · Accepted Answer

Well, the current definition of #"STP"# is.... a temperature of #273.15*K# and a pressure of exactly #10^5*Pa# #(100*kPa, 1*"bar")#.

And #1*"bar"-=(1*"bar")/(1.01325*atm*"bar"^-1)=0.987*atm# I do all this rigmarole BECAUSE I know that #1*atm# will support a column of mercury that is #760*mm# high........... And hence in terms of the length of a mercury column, #1*"bar"-=760*mm*Hg*atm^-1xx0.987*atm=750.0*mm*Hg# #(P_1V_1)/T_1=(P_2V_2)/T_2# We want #V_2=(P_1V_1)/T_1xxT_2/P_2# #(600.0*mm*Hgxx2.02*L)/(303.15*K)xx(273.15*K)/(750*mm*Hg)# #~=1.5*L#

Xavier Avery · Answer

undefined To solve this problem, you can use the combined gas law, which states:

$$ P_1 \times V_1 \div T_1 = P_2 \times V_2 \div T_2 $$

Where:
- $ P_1 $ and $ P_2 $ are the initial and final pressures, respectively.
- $ V_1 $ and $ V_2 $ are the initial and final volumes, respectively.
- $ T_1 $ and $ T_2 $ are the initial and final temperatures, respectively.

Given:
- $ P_1 = 600 \, \text{mm Hg} $
- $ T_1 = 303.15 \, \text{K} $
- $ V_1 = 2.02 \, \text{L} $
- $ P_2 = 750 \, \text{mm Hg} $
- $ T_2 = 273.15 \, \text{K} $

We need to find $ V_2 $, the final volume.

First, we rearrange the equation:

$$ V_2 = \frac{{P_1 \times V_1 \times T_2}}{{P_2 \times T_1}} $$

Then, we plug in the given values and solve for $ V_2 $:

$$ V_2 = \frac{{600 \, \text{mm Hg} \times 2.02 \, \text{L} \times 273.15 \, \text{K}}}{{750 \, \text{mm Hg} \times 303.15 \, \text{K}}} $$

$$ V_2 = \frac{{330453.39}}{{227372.25}} $$

$$ V_2 \approx 1.45 \, \text{L} $$

Therefore, the gas will occupy approximately 1.45 liters at 273.15 K temperature under 750 mm Hg pressure.