If 0.750 L of argon at 1.50 atm and 177°C and 0.235 L of sulfur dioxide at 95.0 kPa and 63.0°C are added to a 1.00-L flask and the flask's temperature is adjusted to 25.0°C, what is the resulting pressure in the flask?
A)0.0851 atm
B)0.244 atm
C)0.946 atm
D)1.74 atm
E)1.86 atm

Question

Quizplus · Accepted Answer

First, we need to convert the temperature of both gases to Kelvin:

177°C + 273.15 = 450.15 K (argon)
63.0°C + 273.15 = 336.15 K (sulfur dioxide)

Next, we need to use the ideal gas law equation to calculate the initial number of moles of each gas:

n(argon) = (P V) / (R T) = (1.50 atm x 0.750 L) / (0.0821 L atm/mol K x 450.15 K) = 0.0344 mol
n(sulfur dioxide) = (P V) / (R T) = (0.0950 MPa x 0.235 L) / (8.31 x 10^-3 L MPa/mol K x 336.15 K) = 0.00994 mol

Now, we can calculate the total number of moles of gas in the flask:

n(total) = 0.0344 mol + 0.00994 mol = 0.0443 mol

Using the total number of moles and the new temperature (25.0°C + 273.15 = 298.15 K), we can solve for the new pressure using the ideal gas law:

P = (n RT) / V = (0.0443 mol x 0.0821 L atm/mol K x 298.15 K) / 1.00 L = 0.946 atm

Therefore, the resulting pressure in the flask is 0.946 atm, so the answer is choice C.

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