A gas has a pressure of 810 kPa at 227°C. What will its pressure be at 27°C?
1 answer:
486 kPa will be the pressure of gas at 27 degrees.
Explanation:
Data given :
Initial pressure of the gas P1 = 810 kPa
initial temperature of the gas T1 = 227 degrees OR 500.15 K
final pressure of the gas =?
final temperature of the gas = 27 degrees OR 300.15 K
Gay Lussac's law is used to calculate the pressure of the gas at 27 degrees or 300.15 K
=
P2 =
Putting the values in the equation:
P2 =
P2 = 486 KPa
486 Kpa is the pressure of the gas when temperature was reduced to 27 degrees.
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Answer: The freezing point of 3.46 gram of a compound X in 160 gram of benzene is
Explanation:
The relation of density and molar mass is:
where
d = density = 3.27 g/ L
P = pressure of the gas = 773 torr = 1.02 atm (760 torr = 1atm)
M = molar mass of the gas = ?
T = temperature of the gas =
R = gas constant =
The relation of depression in freezing point with molality:
= depression in freezing point =
=
= freezing point constant = 5.1
m = molality =
Thus the freezing point of 3.46 gram of a compound X in 160 gram of benzene is