A eudiometer contains 65.0 ml sample of gas collected by the diplacement of water. The water levels inside and outside the eudiometer are the same. The atmospheric pressure is 92.5 kPa and the temperture is 23 degrees. What volume will the dry gas occupy at STP? The vapor pressure of water at 23 degrees is 2.8 kPa.
92.5kPa-2.8kPa= 89.7kPa
65.0 ml(89.7 kPa)/101.3 kPa
answer: 57.6 ml
The only difficulty here is the temp. As I see it you correctly calcuated the volume at standard pressure, 23C. You need to adjust the volume for temperature now.
I think this might be right...adjusting the volume for temperature.
273K + 23K= 296K
The two possible temperature ratio would be 296K/273K or 273K/296K. I chose the second one.
57.6 ml(273K)/296K
answer: 53.1 ml
That looks good to me.
To find the volume of the dry gas at STP, we need to consider the difference in pressure and the ideal gas law.
First, subtract the vapor pressure of water (2.8 kPa) from the atmospheric pressure (92.5 kPa) to get the partial pressure of the dry gas:
92.5 kPa - 2.8 kPa = 89.7 kPa
Next, we can use the equation:
(Volume of wet gas)(partial pressure of dry gas) = (Volume of dry gas)(STP pressure)
Let's substitute the given values into the equation:
(65.0 ml)(89.7 kPa) = (Volume of dry gas)(101.3 kPa)
Now, solve for the volume of the dry gas:
Volume of dry gas = (65.0 ml)(89.7 kPa) / 101.3 kPa
Volume of dry gas ≈ 57.6 ml
Therefore, the volume of the dry gas at STP is approximately 57.6 ml.