at STP 1 mol is 22.4 liters
n = P V/R T
1 = 1*22.4 /R(273)
R = 22.4/273
Now our problem
1 = 1 * 18.2 *273/ (22.4)K
K = 222
222-273 = - 51 deg C
My teacher gave us a packet with many questiona on it for homework. I was able to figure all 44 questions except the following 2.... Help!?!?!
9. "In order to have 1.00 mole of gas fit in a box that measures 1.30 dm x 2.40 dm x 5.83 dm at 1.00 atm, what must the temperature be (in C)? (1L=1dm^3)"
10. "A cube shaped box is to be made that can hold precisely 40.0 grams of He at 1.05 atm and 55C. How long would the box have to be? (remember it's a cube so take the cube root of the formula) "
2 answers
55+ 273 = 328 K
He = 4 g/mol so we have 10 moles
P1V1/(n1 T1) = P2V2/(n2 T2)
22.4/273 = 1.05 V2/(10*328)
V2 = 269 dm^3
side of cube = 6.48 dm
= 64.8 cm or 0.648 m
He = 4 g/mol so we have 10 moles
P1V1/(n1 T1) = P2V2/(n2 T2)
22.4/273 = 1.05 V2/(10*328)
V2 = 269 dm^3
side of cube = 6.48 dm
= 64.8 cm or 0.648 m