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Equal masses of H2, O2 and methane have been taken in a container of volume V at
temperature 27o C in identical conditions. The ratio of the volumes of gases H2:O2:methane
would be :
temperature 27o C in identical conditions. The ratio of the volumes of gases H2:O2:methane
would be :
- a)16 : 1 : 2
- b)8 : 1 : 2
- c)8 : 16 : 1
- d)16 : 8 : 1
Correct answer is option 'A'. Can you explain this answer?
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Given information:
- Equal masses of H2, O2 and methane are taken in a container of volume V
- Temperature = 27°C
To find:
- Ratio of volumes of gases H2:O2:methane
Solution:
1. Calculate the number of moles of each gas using the ideal gas equation: PV = nRT
- P = pressure (assume 1 atm)
- V = volume
- n = number of moles
- R = gas constant
- T = temperature in Kelvin (27°C = 300 K)
For H2:
PV = nRT
(1 atm) V = n (0.0821 L atm/mol K) (300 K)
n(H2) = V/24.8
For O2:
PV = nRT
(1 atm) V = n (0.0821 L atm/mol K) (300 K)
n(O2) = V/49.6
For methane:
PV = nRT
(1 atm) V = n (0.0821 L atm/mol K) (300 K)
n(CH4) = V/22.4
2. Calculate the volume of each gas using the ideal gas equation: PV = nRT
- P = pressure (assume 1 atm)
- V = volume
- n = number of moles calculated in step 1
- R = gas constant
- T = temperature in Kelvin (27°C = 300 K)
For H2:
PV = nRT
(1 atm) V(H2) = (V/24.8) (0.0821 L atm/mol K) (300 K)
For O2:
PV = nRT
(1 atm) V(O2) = (V/49.6) (0.0821 L atm/mol K) (300 K)
For methane:
PV = nRT
(1 atm) V(CH4) = (V/22.4) (0.0821 L atm/mol K) (300 K)
3. Simplify the volume ratios by dividing each volume by the smallest volume
Let x be the smallest volume, then:
- Volume ratio of H2:O2:methane = (V(H2)/x):(V(O2)/x):(V(CH4)/x)
- Divide each volume by x to get the simplified ratio
For H2:
V(H2)/x = [(V/24.8) (0.0821 L atm/mol K) (300 K)]/x
V(H2)/x = (V/19.8)/x
For O2:
V(O2)/x = [(V/49.6) (0.0821 L atm/mol K) (300 K)]/x
V(O2)/x = (V/39.2)/x
For methane:
V(CH4)/x = [(V/22.4) (0.0821 L atm/mol K) (300 K)]/x
V(CH4)/x = (V/17.6)/x
4. Simplify the ratio by dividing each term by the smallest term
- Divide each simplified volume ratio by the smallest volume ratio to get the final ratio
Let y be the smallest term, then:
- Ratio of volumes of H2:O2:methane = (V(H2)/y):(V(O2
- Equal masses of H2, O2 and methane are taken in a container of volume V
- Temperature = 27°C
To find:
- Ratio of volumes of gases H2:O2:methane
Solution:
1. Calculate the number of moles of each gas using the ideal gas equation: PV = nRT
- P = pressure (assume 1 atm)
- V = volume
- n = number of moles
- R = gas constant
- T = temperature in Kelvin (27°C = 300 K)
For H2:
PV = nRT
(1 atm) V = n (0.0821 L atm/mol K) (300 K)
n(H2) = V/24.8
For O2:
PV = nRT
(1 atm) V = n (0.0821 L atm/mol K) (300 K)
n(O2) = V/49.6
For methane:
PV = nRT
(1 atm) V = n (0.0821 L atm/mol K) (300 K)
n(CH4) = V/22.4
2. Calculate the volume of each gas using the ideal gas equation: PV = nRT
- P = pressure (assume 1 atm)
- V = volume
- n = number of moles calculated in step 1
- R = gas constant
- T = temperature in Kelvin (27°C = 300 K)
For H2:
PV = nRT
(1 atm) V(H2) = (V/24.8) (0.0821 L atm/mol K) (300 K)
For O2:
PV = nRT
(1 atm) V(O2) = (V/49.6) (0.0821 L atm/mol K) (300 K)
For methane:
PV = nRT
(1 atm) V(CH4) = (V/22.4) (0.0821 L atm/mol K) (300 K)
3. Simplify the volume ratios by dividing each volume by the smallest volume
Let x be the smallest volume, then:
- Volume ratio of H2:O2:methane = (V(H2)/x):(V(O2)/x):(V(CH4)/x)
- Divide each volume by x to get the simplified ratio
For H2:
V(H2)/x = [(V/24.8) (0.0821 L atm/mol K) (300 K)]/x
V(H2)/x = (V/19.8)/x
For O2:
V(O2)/x = [(V/49.6) (0.0821 L atm/mol K) (300 K)]/x
V(O2)/x = (V/39.2)/x
For methane:
V(CH4)/x = [(V/22.4) (0.0821 L atm/mol K) (300 K)]/x
V(CH4)/x = (V/17.6)/x
4. Simplify the ratio by dividing each term by the smallest term
- Divide each simplified volume ratio by the smallest volume ratio to get the final ratio
Let y be the smallest term, then:
- Ratio of volumes of H2:O2:methane = (V(H2)/y):(V(O2
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Equal masses of H2, O2 and methane have been taken in a container of volume V attemperature 27o C in identical conditions. The ratio of the volumes of gases H2:O2:methanewould be :a)16 : 1 : 2b)8 : 1 : 2c)8 : 16 : 1d)16 : 8 : 1Correct answer is option 'A'. Can you explain this answer?
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