What is the volume occupied by 0.108 mol of helium gas at a pressure of 0.909 atm and a temperature of 306 K ? Express your answer using three significant figures. Would the volume be different if the gas was argon (under the same conditions)? The volume would be the lower for argon gas. The volume would be the same for argon gas. The volume would be the greater for argon gas.
Answers
Answer: 2.93 L
Explanation:
AI-generated answer
To find the volume occupied by 0.108 mol of helium gas at a pressure of 0.909 atm and a temperature of 306 K, we can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Rearranging the equation to solve for V:
V = (nRT) / P
Plugging in the given values:
n = 0.108 mol
R = 0.0821 L·atm/(mol·K) (the ideal gas constant)
T = 306 K
P = 0.909 atm
V = (0.108 mol * 0.0821 L·atm/(mol·K) * 306 K) / 0.909 atm
Calculating this expression, we find that the volume occupied by 0.108 mol of helium gas at a pressure of 0.909 atm and a temperature of 306 K is approximately 2.93 L.
Now, let's consider the second part of the question: Would the volume be different if the gas was argon (under the same conditions)?
The volume would be the same for argon gas.
According to the ideal gas law, at the same temperature, pressure, and number of moles, the volume occupied by a gas is the same regardless of the gas's identity. Therefore, if we replaced helium gas with argon gas while keeping the same conditions of pressure, temperature, and number of moles, the volume occupied by argon gas would be the same, approximately 2.93 L.
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Answers
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Explanation:
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Answers
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