when 2g of a gas A is introduced into an evacuated flask kept at
25 centigrade, the pressure is found to be 1 atmosphere.if 3g of another gas is then added to the same flask, the total pressure becomes 1.5atm. assuming ideal behaviour, the ratio of their molecular weights is
call the 2nd gas B
3 g of B is half as many moles as 2 g of A
B / A = 3 / 1
To determine the ratio of molecular weights of the two gases, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L*atm/mol*K)
T = temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
Given that the temperature is 25°C, we obtain:
T(K) = 25 + 273.15 = 298.15K
Now, let's calculate the number of moles of gas A using its mass (2g) and molar mass (M(A)):
n(A) = mass/M(A)
Similarly, for gas B, we have:
n(B) = mass/M(B)
Initially, when 2g of gas A is introduced into the flask, the pressure is 1 atmosphere. Thus, we can write:
PV = n(A)RT
1 * V = n(A) * R * 298.15
Now, when an additional 3g of gas B is added to the flask, resulting in a total pressure of 1.5 atmospheres, we can write:
P'V = (n(A) + n(B))RT
1.5 * V = (n(A) + n(B)) * R * 298.15
Next, we can divide the second equation by the first equation to eliminate the volume:
(1.5 * V) / (1 * V) = [(n(A) + n(B)) * R * 298.15] / [n(A) * R * 298.15]
1.5 = (n(A) + n(B)) / n(A)
Now, substitute the values for n(A) and n(B) in terms of mass and molar masses:
1.5 = (2 / M(A) + 3 / M(B)) / (2 / M(A))
To simplify further, we can multiply both sides by M(A):
1.5 * M(A) = (2 / M(A) + 3 / M(B))
Now, cross-multiply:
1.5 * M(A) * (2 / M(A) + 3 / M(B)) = 2
Simplify the left side:
3 + (4.5 * M(A) / M(B)) = 2
Rearrange the equation:
4.5 * M(A) / M(B) = -1
Finally, solve for the ratio of molecular weights (M(A) / M(B)):
M(A) / M(B) = -1 / 4.5 = -2 / 9
The ratio of molecular weights is -2/9.