chemistry question :(?

Question

16. at 525 mm Hg and 85 degrees C, the volume of a sample of nitrogen gas is 26.8 L. what is the volume at STP?


A. 1 L
B. 10.9 L
C. 15.3 L
D. 15.3 L c, and d are same it shows on the test?...

Answer 1

PV = nRT

P = 525 mmHg
V = 26.8 L
R = 62.3637 L • mmHg / K • mole
T = 86ºC + 273.15 = 359.15 K

n = PV/RT = 525*26.8/(62.36)(359.16) = 0.628 moles

at STP

T = 298 K
P = 760 mmHg
V = ?
R = 62.3637 L • mmHg / K • mole
n = 0.628 moles

V = 0.628*62.36*298/(760) = 15.35 L

so it could be either C or D...?

Answer 2

At STP
760 mm Hg (101 kPa)
25 degrees C (298 K)

Using the 'Ideal Gas Eq'n' PV = nRT
n = PV/RT
So initially
nR = 69.77 x 0.0268 /358 = 101 x V /198
Eliminating 'nR'
V = (69.77 x 0.0268 x 198) /(358 x 101) = 0.0102 m^3
V = 10.239 L

NB When using the Ideal Gas Equation:-
Pressure in Pascals 525 becomes 69.77 and 760 becomes 101.
Volume in Cubic Metres(m^3) 26.8L = 0.268 m^3
Temperature in Kelvin 85 = 358 and 25 = 198
Answer in Cubic metres and multiply by 1000 as 1000 litres makes 1 m^3.

Answer 3

Just remember the ideal gas law PV=nRT
"n" doesn't change in this problem...neither does "R"; therefore-
PV/T = pv/t
0.691 atm x 26.8 L / 358.15 deg K = 1 atm x L / 273.15 deg K
0.05169 = liters / 273.15
liters = 14.1 liters
correct answer is 14.1 liters