Consider three 1.0 L flasks at 25° C and 96.66 kPa containing the gases CH4 (flask A), CO2 (flask B), and C2H6 (flask C). In which flask is there 0.039 moles of gas?

flask A
flask C
flask B
none
all

Use PV=nRT where n=.039 and see which one balances.

To determine which flask contains 0.039 moles of gas, we need to compare the number of moles of gas in each flask.

First, let's calculate the number of moles of gas in each flask using the Ideal Gas Law equation:

PV = nRT

Where:
P = pressure (in kilopascals, kPa)
V = volume (in liters, L)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K) or 8.314 J/(mol·K))
T = temperature (in Kelvin, K)

Given:
For flask A:
P_A = 96.66 kPa
V_A = 1.0 L
T_A = 25°C = 25 + 273.15 K

For flask B:
P_B = 96.66 kPa
V_B = 1.0 L
T_B = 25°C = 25 + 273.15 K

For flask C:
P_C = 96.66 kPa
V_C = 1.0 L
T_C = 25°C = 25 + 273.15 K

Now we can calculate the number of moles of gas in each flask using the rearranged Ideal Gas Law equation:

n = PV / RT

For flask A:
n_A = (P_A * V_A) / (R * T_A)

For flask B:
n_B = (P_B * V_B) / (R * T_B)

For flask C:
n_C = (P_C * V_C) / (R * T_C)

Substituting the given values into the equations, we can calculate the number of moles of gas in each flask.

n_A = (96.66 kPa * 1.0 L) / (0.0821 L·atm/(mol·K) * (25 + 273.15) K)

n_B = (96.66 kPa * 1.0 L) / (0.0821 L·atm/(mol·K) * (25 + 273.15) K)

n_C = (96.66 kPa * 1.0 L) / (0.0821 L·atm/(mol·K) * (25 + 273.15) K)

Calculating n_A, n_B, and n_C, we can compare the values to find the flask containing 0.039 moles of gas.