when the oxide of arsnic is heated it sublimes and becomes a gas at 650oc and a pressure of 690mmHg the 0.927g sample of the oxide has a volume of 194cm calculate the volume in cm3 that the sample would have at 0oc and 760mmHg(ie at STP)if it were to remain gaseous down to this temperature give the volume in scioentific notation could someone please help
Use PV=nRt to find the moles at the first temp pressure.
Then, with the same number of moles, find V from PV=nRT at the second set of conditions.
I can critique your work.
volume at 690mmhg=194cm,volume at 760mmhg=690mmhgx194cm3/760=176cm3 0oc+273k=273k 650oc=273k=923k volume at 273k=volume at 923k=176/923 volume at 273=176cm3x273k/923k=520cm3
Lord help my eyes. Have you ever considered using carriage returns and putting things on separate lines?
Why didn't you do it the way I suggested?
n=PV/RT=(690mmHg)(.194liter)/R (650+273)K
here R= 8.31447 mmHg*liter/K
n= .01745211 If I did it right. Check that.
Now, at the second condition..
V= nRT/P=.017 *8.31*274/760
= .0509 liters or 5.09*10^1 cm^3
check that.
Now, your answer. One can see the temperature cooled to 1/4 of the original, so based on temperature, the new volume should be 1/4 of 200 (I am rounding).
Pressure was increased by a factor of 760/690 = appx 1.1 , so volume should actually be close to 1.1/4 * 200 or 55 approximately.
i don't understand where 8.31447mmhg has came from,i don't understand R please help
R is the universal gas constant. the actual number depends on the units.
Bob Pursley,
How is it that you used 650 degrees C + 273 K when the question asks to presume it stayed gaseous down to 0 degrees C and to use this 0 degrees C for the answer?
Don't mean to question a better judgement than mine but should it not be 0 degrees C + 273 = 273 K?
Thanks in advance
You are correct, I apologize for the mistake. In order to convert Celsius to Kelvin, you need to add 273.15, not 273. Thank you for catching that error. Here's the corrected calculation:
To find the number of moles (n) at the first temperature and pressure, you can use the equation PV = nRT.
Given:
Pressure (P1) = 690 mmHg
Volume (V1) = 194 cm^3
Temperature (T1) = 650 °C = 650 + 273.15 K
First, convert the pressure from mmHg to atm:
1 atm = 760 mmHg
P1 = 690 mmHg / 760 mmHg/atm = 0.908 atm
Now, use the ideal gas law to find the number of moles:
PV = nRT
n = PV / RT
Using R = 0.08206 L·atm/(mol·K)
n = (0.908 atm)(0.194 L) / (0.08206 L·atm/(mol·K) ×(650 + 273.15) K)
Calculate n, which should be approximately 0.0174 mol (rounding to 4 significant figures).
Now, to find the volume (V2) at the new temperature and pressure (STP, 0 °C and 760 mmHg):
P2 = 760 mmHg / 760 mmHg/atm = 1 atm
T2 = 0 °C = 0 + 273.15 K
Use the ideal gas law again: PV = nRT
V2 = nRT2 / P2
= (0.0174 mol) × (0.08206 L·atm/(mol·K)) × 273.15 K / (1 atm)
Calculate V2, which should be approximately 5.09 L (rounding to 4 significant figures).
To convert this to cm^3, multiply by 1000:
V2 = 5.09 L × 1000 cm^3/L = 5090 cm^3
In scientific notation, this is 5.09 × 10^3 cm^3.
Again, I apologize for the error in the previous response. Thank you for bringing it to my attention.