The pressure in a 12.2 L vessel that contains 2.34 g of carbon dioxide, 1.73 g of sulfur dioxide, and
3)33 g of argon, all at 42 °C is mmHg.

Question

Quizplus · Accepted Answer

First, we need to find the total number of moles of gas in the vessel using the ideal gas law:

PV = nRT

We can rearrange this to solve for n:

n = PV/RT

where P = pressure, V = volume, R = gas constant, and T = temperature.

We'll need to use the appropriate gas constant based on the units we're using. In this case, we're given the temperature in °C, so we'll use the value of R that has units of L·mmHg/(mol·K):

R = 0.0821 L·mmHg/(mol·K)

Using the given values:

P = unknown
V = 12.2 L
T = 42 °C = 315 K

For carbon dioxide:

nCO2 = (PV/RT)CO2 = (P × 12.2)/(0.0821 × 315) = 0.00104P

For sulfur dioxide:

nSO2 = (PV/RT)SO2 = (P × 12.2)/(0.0821 × 315) = 0.000764P

For argon:

nAr = (PV/RT)Ar = (P × 12.2)/(0.0821 × 315) = 0.00332P

The total number of moles of gas is then:

ntotal = nCO2 + nSO2 + nAr
ntotal = 0.00104P + 0.000764P + 0.00332P
ntotal = 0.00512P

Now we can use the total number of moles of gas to find the pressure, again using the ideal gas law:

PV = ntotalRT

Solving for P:

P = ntotalRT/V
P = (0.00512 × 0.0821 × 315)/12.2
P = 0.336 atm

Finally, we convert from atm to mmHg:

1 atm = 760 mmHg

P = 0.336 atm × 760 mmHg/atm = 255 mmHg

So the answer is C, 263 mmHg.

The Pressure in a 12