Question #86557
At 293 mm Hg and 47.5 °C, a sample of carbon monoxide gas occupies a volume of 0.243 L. The gas is transferred to a 1.13 L flask and the temperature is reduced to 6.56 °C. What is the pressure of the gas in the flask?

A. 72.2 mm Hg
B. 8.70 mm Hg
C. 7.23×10-2 mm Hg
D. 1.19×103 mm Hg
E. 55.0 mm Hg
1
Expert's answer
2019-03-22T04:27:56-0400

Clapeyron-Mendeleev equation for ideal gas is

PV=mMRTPV= \frac{m}{M} RT,

where P – gas pressure; V – gas volume; m and M - mass and molar mass of the gas, respectively; R – gas constant, T – gas temperature. m and M are constant under the considered conditions, so PVT=const\frac{PV}{T}=const.

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}

P2=P1V1T2T1V2P_2=\frac{P_1 V_1 T_2}{T_1 V_2}

where P1 and P2 – initial and final gas pressures, respectively; V1 and V2 – initial and final gas volumes, respectively; T1 and T2 – initial and final gas temperatures, respectively.

T[K]=T[°C]+273.15T [K] = T [\degree C] +273.15

T1=47.5+273.15=320.65  KT_1=47.5+273.15=320.65 \; K

T2=6.56+273.15=279.71  KT_2=6.56+273.15=279.71 \; K

P2=293[mm  Hg]0.243[L]279.71[K]320.65[K]1.13[L]55.0  mm  HgP_2=\frac{293 [mm \; Hg] \cdot 0.243 [L] \cdot 279.71 [K]}{320.65 [K] \cdot 1.13 [L]} \approx 55.0 \; mm \; Hg


Answer: E. 55.0 mm Hg 


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