Give the half-equation for the reaction at the anode, during
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Answer
A)The volume decreases by a factor of 4
B), the volume has increased by factor of 2
Explanation:
A)Given:
P1= 760Kpa
P2 =202Kpa
The temperature changes from37C to155C
There is increase In pressure from P1 to P2
P1= 760torr.
We need to convert to Kpa
But, 1atm= 760torr
Then 760torr 101000pa
Then 101000pa = 101Kpa
We need to convert the temperature from Celsius to Kelvin
T1= 37+273= 310K
But from ideal gas, we know that PV = nRT where nR is constant
Where P= pressure
V= volume
T= temperature
n = number of moles
(P1V1/T1)=(P2V2/T2)
V1/V2 = P2/P1 * T1/T2
V1/V2 = (202/101)*(310/155)
V1/V2=4
Therefore, the volume has decreased by factor of 4
B)
Given:
P1= 2atm
P2 =101Kpa
The temperature changes from 305K to 32C
There is increase In pressure from P1 to P2
P1= 2atm
We need to convert to Kpa
But, 1atm= 760torr
Then 760torr 101000pa
Then 101000pa = 101Kpa
P1= 202.65kpa
We need to convert the temperature from Celsius to Kelvin
T2= 32+273= 305K
But from ideal gas, we know that PV = nRT
Where P= pressure
V= volume
T= temperature
n = number of moles
(P1V1/T1)=(P2V2/T2)
V1/V2 = P2/P1 * T1/T2
V1/V2 = (202/101)
V1/V2 = (101/202.65)*(305/305)
V1/V2 = 1/2
Therefore, the volume has increased by factor of 2