A sample of gas weighing 9.0 (g) at a pressure of 1 atm occupies a volume of 12.3 L. If the pressure is doubled,what is the resulting volume? (Assume constant temperature)

P doubled, volume is halved.

To solve this problem, we can use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.

Since we are assuming constant temperature, we can rewrite the equation as:
P₁V₁ = P₂V₂

Let's calculate the initial number of moles of gas using the given information.
m = mass of gas = 9.0 g
M = molar mass of gas (unknown in this case)

To find the molar mass (M), we can use the equation:
M = (m/M) × RT/P (where R = ideal gas constant)

R = 0.0821 L·atm/(mol·K) (given)
T = temperature (constant, not given)
P = pressure at the start = 1 atm

Now, we can substitute the given values into the equation to solve for the molar mass (M). However, since the molar mass of the gas is not given in the question, we cannot proceed further to find the initial number of moles.

Hence, without the molar mass, it is not possible to calculate the resulting volume when the pressure is doubled.