At 0.00 °C and 1.00 atm pressure, a sample of gas occupies 20.0 mL. If the temperature is increased to 30.0°C and the entire gas sample is transferred to a 30.0-mL container. The pressure of the gas in the new container is:
• 0.499 atm
• 0.740 atm
• 1.35 atm
• 1.66 atm
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature in Kelvin
First, we need to convert the temperatures from Celsius to Kelvin:
0.00 °C = 273.15 K
30.0 °C = 303.15 K
Next, we can calculate the initial number of moles using the initial conditions:
P1V1 / T1 = nR
(1.00 atm)(20.0 mL) / (273.15 K) = n(0.08206 L.atm/mol.K)
n = 8.00 x 10^-4 mol
Now, we can use the ideal gas law again to calculate the pressure of the gas in the new container:
P2 = nRT2 / V2
P2 = (8.00 x 10^-4 mol)(0.08206 L.atm/mol.K)(303.15 K) / 30.0 mL
P2 = 0.499 atm
So, the pressure of the gas in the new container is 0.499 atm. The correct answer is:
• 0.499 atm