molar mass, atm, pressure, find mass. HELPby teggenspiller Tags: kinetic energy, nitrogen, thermodynamics 

#1
Jun811, 12:30 PM

P: 94

1. The problem statement, all variables and given/known data
A nitrogen in the container (molar mass is 28 g/mol) has a pressure of 2.5 atm, a volume of 0.2 m3, and a temperature of 300C. What is the mass of the nitrogen? 2. Relevant equations first figure how many moles there are n=PV/RT however, i need to use Pressure, Pa, and not in atmospheres. 3. The attempt at a solution how do i figure the pressure in Pa, and not atm. or can I use atm? hmmmm 



#2
Jun811, 01:22 PM

Mentor
P: 11,421

Google: 1 atm in Pascals




#3
Jun811, 01:25 PM

P: 94

i did its 101.3 or 103.1. but either way i couldnt come up with correct answer... :( can u show me how to solve this.. ?




#4
Jun811, 01:28 PM

P: 94

molar mass, atm, pressure, find mass. HELP
VP/RT
.2*253312.5/(8.31*303)= # of moles, multiplied by 28 grams per mole *(28) 563.37944 



#5
Jun811, 01:32 PM

Mentor
P: 11,421

Notes: 1. 1 atm = 101,325 pascals or 101.3 kPa 2. Make sure that your temperature is absolute (Kelvin). 



#6
Jun811, 01:35 PM

P: 94

okay thank you. one minute.




#7
Jun811, 01:40 PM

P: 94

P= (253312.5) Pa
R=8.31 J/(mol K) V= in Liters of m^3?? V= .2m^3 or 200Liters T= 273.15+30 = 303.15 (253312.5) * ( .2) / (8.31J/(mol K) *(303.15)= 1848175.32 *28g/mol =51748908.8 much too big 



#8
Jun811, 01:49 PM

Mentor
P: 11,421

I thought that your temperature was 300C? You've used 30
The formula is n = p*v/(R*T) ; you have to divide by temperature, not multiply! 



#9
Jun811, 01:50 PM

P: 94

ha okay




#10
Jun811, 01:50 PM

P: 94

jeez i completely missed that. thnaks!




#11
Jun811, 02:18 PM

P: 94

i still am not geting the right thing? do i times the final answer by the grams/mol (28//)




#12
Jun811, 02:20 PM

Mentor
P: 11,421





#13
Jun811, 02:32 PM

P: 94

A. 125 g
B. 258 g C. 421 g D. 582 g E. 864 g 



#14
Jun811, 02:56 PM

Mentor
P: 11,421

What final value for the grams of N2 did you calculate? Is it possible that the starting values were modified to 'present a new problem'? 


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