Answer:
Note: always used kelvin temperature in computing gas law problems. K = Degree celsius + 273.15 (sometimes your teacher will 273 instead of 273.15 .
1. Given:
T1 = 373.15 K (K= degree celsius + 27.15) K = 100+273.15 = 373.15 K
T2 = 473.15 K (K= degree celsius + 27.15) K = 200+273.15 = 473.15 K
V1 = 8 L
V2 = ? unknown
Solution:
V2 = V1 x T2 / T1
= 8L × 473.15 K / 373.15 K
= 10.14 L
2. Given
V1 = 20 ml
T2 = 373.15 (K= degree celsius + 27.15)
K = 100 + 273.15 = 373.15 K
T1 = 330 K
V2 = ?
Solution:
V2 = V1 x T2 / T1
= 20 ml × 373.15 / 330 K
= 22.615 ml
3. Give:
P = 3.0 atm
V = 7 L
T = 310.15 K (K= degree celsius + 27.15)
K = 37 + 273.15 = 310.15 K
n= ? unknown
Solution:
PV = nRT
n = PV/RT
= 3.0 × 7 / 0.0821 × 310.15
= 21 / 25.4633
= 0.825 mol
you can solve for the grams by multiplying 0.825 to the molar mass of the given gas.
Explanation:
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