A 15.0 L sample of CO gas initially has a temperature of 23.5oC at a pressure of 1.16 atm. What is the new pre?
A 15.0 L sample of CO gas initially has a temperature of 23.5oC at a pressure of 1.16 atm. What is the new pressure, in atm, if the volume is changed to 12.5 L and the temperature is decreased to 18.7oC?
- 10 years agoFavorite Answer
The standard equation for finding the solutions is PV/NT = PV/NT and number of moles stays the same so
1.16 x 15 / 296.5( T in Kelvin) = X x 12.5 / 291.7.
solve to find your new pressure in ATM is 1.369465
Correct sig figs shows 1.37
- Anonymous10 years ago
Use the Combined Gas Law that looks like this:
(P1 V1)/ (T1) = (P2 V2) /( T2)
P1= 1.16 atm
V1= 15.0 L
T1= 23.5 degrees celsius. For all Gas Law problems Celsius has to be converted to Kelvin by adding 273.15 K to the given celsius temperature
So 23.5 celsius + 273.15 K = 296.65 K
P2= ? (we have to find this)
V2= 12.5 L
T2= 18.7 degrees celsius + 273.15 K =291.85 K
Now plug in these values into the formula
(1.16 atm) * (15.0L) / (296.65 K) = (P2) * (12.5 L) / (291.85 K)
Cross multiply so u multiply 1.16 * 15.0 * 291.85 and divide by 296.65 * 12.5
This will give u your new pressure or P2 = 1.37 atmSource(s): Pre Med student
- buffingtonLv 44 years ago
There are different strategies that is used... extremely of the mixed or appropriate gas regulation, i will use here 2 rules. a million...Boyle's regulation: P1xV1 = P2xV2. 3.5atm x 40 seven.3L = 2.05atm x V2. V2 = (3.5 x 40 seven.3) / 2.05 = 80.8L. 2...Charles' regulation. V1xT2 = V2xT1 (27.5°C + 273 = 3 hundred.5K = T2 (35°C + 273 = 308K = T1.) 80.7L x 3 hundred.5K = V2 x 308K V2 = (80.8 x 3 hundred.5) / 308 = seventy 8.8L = answer a)...