chemistry problem?

Question

A gas has volume of 20l measured at 27degreescelsius and 4 atmospheres. what will be its volume at STP?

Answer 1

v1p1=v2p2[v1=the volume at 1 atmos pressure and 27c
v2=v at 4 atm p 27c p1=normal p=1 atm p p2=4at

-- v1*1=20l * 4
v1=80l
acc to charles law
v1=v0(1+t/273)[v0=v at stp]
80=vo(1+27/273)
80=v0* 300/273
v0=80*273/300
v0=72.8l

Answer 2

Remember, at STP, pressure = 100.00 kPa or 1 bar and temperature = 273.15 K (or 0 degrees celsius)

Using the ideal gas law:

(P1) x (V1) / (T1) = (P2) x (V2) / (T2)

From the given info:

P1 = 4 atmospheres (convert to bar or kPa)
V1 = 20 liters
T1 = 27 degrees celsius
P2 = 100 kPa or 1 bar (make sure if you use kPa for P1, use kPa for P2, and so goes for bar)
T2 = 0 degrees celsius

Rearrange the equation above to solve for V2 (volume at STP)

Answer 3

conditions STP T0= 273K p0=10^5 Pa

use pV=nRT so p0V0 =nRT0 (0) and p1V1=nRT1(1)

divide (0) by (1)

p0V0/p1V1=T0/T1

and V0=p1V1T0/p0T1 = 4*101325*20*273/100000*300=73.76L

Answer 4

27+273 = 300K
P1V1/T1 = P2V2/T2
4x20/300 = 1xV2/273
V2 = 72.8 liters