Question

If 200. mL of N2 at 25°C and pressure 400. mm Hg, and 200. mL of O2 and at 25° and pressure of 300. mm Hg are placed in a 700. mL container, what is the total pressure?

Answer 1

For N2, volume, V=200 mL=0.2 L Temperature, T=25°C=25+273=298 K, Pressure, P=400 mmHg.

From ideal gas equation, PV=nRT

Where R=62.3636 L mmHg mol^-1 K^-1 and n=number of moles.

Therefore n(N2)=PV/RT=(400 mmHg x 0.2 L)/(62.3636 L mmHg mol^-1 K^-1 x 298 K)

n=0.0043 mol of N2.

Similarly for O2, V=200 mL=02 L, P=300 mmHg, T=25°C=298 K

Therefore moles of O2,

n(O2)=(300 mmHg x 0.2 L)/(62.3636 L atm mol^-1 K^-1 x 298 K)

n=0.00323 mol of O2.

Now volume=700 mL=0.7 L, T=298 K, n=n(N2) + n(O2)

n=0.0043+0.00323=000753 mol

Therefore total pressure, P=nRT/V

P=(0.00753 mol x 62.3636 mmHg x 298 K)/(0.7 L)

P=199.87 mmHg ~ 200 mmHg.

Please let me know if you have any doubt. Thanks