How many grams of butane were in 1. 000 atm of gas at room temperature?
1 answer:
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Answer:
1223.38 mmHg
Explanation:
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value =
Also,
Moles = mass (m) / Molar mass (M)
Density (d) = Mass (m) / Volume (V)
So, the ideal gas equation can be written as:
Given that:-
d = 1.80 g/L
Temperature = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (32 + 273.15) K = 305.15 K
Molar mass of nitrogen gas = 28 g/mol
Applying the equation as:
P × 28 g/mol = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K
⇒P = 1223.38 mmHg
<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>