the pressure on one mole of a gas at STP is doubled and the temperature is raised to 546 K what is the final volume of gas when 1 mole of gas occupies 22.4 DM cube at STP
Answers
Answered by
51
Given conditions are-
At STP, volume of gas is (V1) = 22.4 dm3
Pressure is (P1) = 1 atm
Temperature is (T1) = 273 K
So, Volume of gas when pressure is doubled (P2) = 2 atm
Temperature is raised to (T2) = 546 K
Applying gas equation
P1V1 / T1 = P2V2 / T2
(1 X 22.4) / 273 = (2 X V2) / 546
0.08 = 0.0036 X V2
V2 = 22.22 dm3
At STP, volume of gas is (V1) = 22.4 dm3
Pressure is (P1) = 1 atm
Temperature is (T1) = 273 K
So, Volume of gas when pressure is doubled (P2) = 2 atm
Temperature is raised to (T2) = 546 K
Applying gas equation
P1V1 / T1 = P2V2 / T2
(1 X 22.4) / 273 = (2 X V2) / 546
0.08 = 0.0036 X V2
V2 = 22.22 dm3
Answered by
14
➰➰hey buddy here's ur answer➰➰
solution➖➖
given:
p1=760mm Hg
T1= 343k
V1=22.4L
, P2= 760×2= 1520mm Hg
T2= 546k
V2=??
then we know that,,
P1V1/T1= P2V2/T2
760×22.4÷343= 1520×V2/545
V2= 546×22.4×2/343
V2= 78×11.2÷49
V2= 873.6/49= 17.82Litres
❤❤hope it helps u❤❤
solution➖➖
given:
p1=760mm Hg
T1= 343k
V1=22.4L
, P2= 760×2= 1520mm Hg
T2= 546k
V2=??
then we know that,,
P1V1/T1= P2V2/T2
760×22.4÷343= 1520×V2/545
V2= 546×22.4×2/343
V2= 78×11.2÷49
V2= 873.6/49= 17.82Litres
❤❤hope it helps u❤❤
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