If I initially have a gas at a pressure of 13.5 atm, a volume of 2300 mL, and a temperature of 200 K, and then I raise the pressure to 14 atm and increase the temperature to 300 K, what is the new volume of the gas? Report the answer in correct number of significant figures and units.
To solve this problem, we will use the ideal gas law, which is given by the formula:
PV = nRT
Where:
P is the pressure of the gas,
V is the volume of the gas,
n is the number of moles of the gas,
R is the ideal gas constant,
T is the temperature of the gas.
First, let's calculate the initial number of moles (n1) using the initial conditions given:
n1 = (PV)/(RT)
P1 = 13.5 atm
V1 = 2300 mL = 2.3 L
T1 = 200 K
R = 0.0821 L·atm/(mol·K)
n1 = (13.5 atm * 2.3 L) / (0.0821 L·atm/(mol·K) * 200 K)
n1 ≈ 1.97 mol
Now, for the new conditions, we need to find the new volume (V2). We can rearrange the ideal gas law equation to solve for V:
V = (nRT)/P
n = n1 = 1.97 mol
R = 0.0821 L·atm/(mol·K)
P = 14 atm
T = 300 K
V2 = (1.97 mol * 0.0821 L·atm/(mol·K) * 300 K) / 14 atm
V2 ≈ 8.56 L
Therefore, the new volume of the gas is approximately 8.56 L. The number of significant figures in the answer is limited by the number of significant figures in the given values, which is 4 (P2, T2).