What is the volume occupied by 0.118mol of Helium gas at a pressure of 0.97atm and a temperature of 305k? Would the volume be different if the gas was Argon(under the same identical conditions)?

To find the volume occupied by a given quantity of gas, we can use the ideal gas law equation:

PV = nRT,

Where:
P = pressure (in atmospheres),
V = volume (in liters),
n = number of moles,
R = the ideal gas constant (0.0821 L.atm/mol·K),
T = temperature (in Kelvin).

We are given:
n (number of moles) = 0.118,
P (pressure) = 0.97 atm,
T (temperature) = 305 K.

We can rearrange the ideal gas law equation to solve for V:

V = (nRT) / P.

Now, we can substitute the given values into the equation to calculate the volume occupied by the helium gas:

V(helium) = (0.118 mol * 0.0821 L.atm/mol·K * 305 K) / 0.97 atm.

Calculating this equation gives us the volume occupied by the helium gas.

To find the volume occupied by argon under the same conditions, we can repeat the same calculation, but use the number of moles of argon:

n (number of moles) = 0.118.

V(argon) = (0.118 mol * 0.0821 L.atm/mol·K * 305 K) / 0.97 atm.

Comparing the two volume values will let us determine if the volume would be different for argon under the same identical conditions.