A sample of sulfur dioxide (SO2) is initially at a temperature of 133 C a volume of 20 L, and a pressure of 850 mm Hg. If the volume changes to 25 L and the temperature increases to 181 C, what is the new pressure?

Show your work. Don't forget to convert C to kelvin

760.39 mmHg (approx.)

Explanation:

Okay so given the first set of information, we have:

T = 406 K (133+273)

V = 20 L

P = 1.1184 atm (converted for continuity reasons from mmHg)

R = 0.08206 L atm/mol K

Using PV = nRT (ideal gas law), we solve for n (moles of SO2).

n = PV/RT = (1.1184 atm * 20 L) / (0.08206 L atm/mol K * 406 K) = 0.6714 mol

Now that you have all the information, you can use the new volume and temperature to solve for pressure.

T = 454 K (181+273)

V = 25 L

n = 0.6714 mol

R = 0.08206 L atm/mol K

Using the ideal gas law again, we solve for P (pressure in atm).

P = nRT/V = (0.6714 mol * 0.08206 L atm/mol K * 454 K) / (25 L) = 1.0005 atm

Then we convert atm back to mmHg to get 760.39 mmHg.