The vapor pressure of carbonyl sulfide (OCS) is given by the following empirical

Question

The vapor pressure of carbonyl sulfide (OCS) is given by the following empirical formula (for 162-224K):

log10(P0/torr) = (-1318.260/T) + 10.15309 - (1.4778x10-2)T+(1.8838x10-5)T2

a) Determine the vapor pressure and heat of vaporization of this substance at 200K

1. The vapor pressure of carbonyl sulfide (ocs is given by the following empirical formula (for 162-224 K) log10 (P Orr 1318.260 +10.15309-(1.4778x 10 2)T+ (1.8838x 10 5)T F (a) Determine the vapor pressure and heat of vaporization of this substance at 200 K. (b) Calculate the temperature for which po 50 torr.

Explanation / Answer

The vapor pressure is log10(P/torr)= -1318.260/T +10.15309- 1.4778*10-2T +1.8838*10-5 T2   (1)

At 200 K, log 10 ( P/torr)= -1318.260/200 +10.15309- 1.4778*10-2*200 +1.8838*10-5*(200)2 =1.35971

Hence, P(Torr)= 22.97

Differentiating the equation , (1/P)*ln10 (dP/dT) = 1318.260/T2 -1.4778*10-2 +2*1.8838*10-5T

At 180K, the vapor pressure log10(P/Torr)= -1318.260/180+10.15309-1.4778*10-2*180+1.8838*10-5*180*180

P(torr)= 6.04 Torr

since d(lnP)= deltaH/RT2

lnP:= -deltaH/RT + C

2.303*logP= -deltaH/RT+C

logP= -deltaH/RT*2.303 +C’ where C’= C/2.303

Comparing the coefficients of 1/T gives from the vapor pressure equation

– deltaH/2.303*R =-1318.260

deltaH= 1318.260*8.314*2.303 = 25240.914 J/moles=25.240 Kj/mole.

since the given equation for logP is non linear, the calculations are done in excel and result is shown below

so the temperature corresponding to 50.00 torr is 214.21 K

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