Question

what is the pressure in atm exerted by 2.48 moles of a gas in a 250.0 ml container at 58c?

Answer 1

T = 58 + 273.15 = 331.15

R = 0.082 atm

V = 250.0 mL / 1000 = 0.25 L

P =  n x R x T / V

P = 2.48 x 0.082 x 331.15 / 0.25

P = 67.342 / 0.25

P = 269.368 atm

hope this helps!

Answer 2

Explanation:

The given data is as follows.

     V = 250.0 ml,                  T = 58^{o}C = 58 + 273 = 331 K

      n = 2.48 moles,              P = ?    ,             R = 0.082 atm L /K mol

Therefore, calculate the pressure using ideal gas equation as follows.

                       PV = nRT

                        P = $\frac{nRT}{V}$

                            = $\frac{2.48 mol \times 0.082 atm L/K mol \times 331 K}{250.0 ml}$

                            = 0.269 atm

Thus, we can conclude that pressure of the given gas is 0.269 atm.