Question

Calculate pressure exerted by a 0.25 mole sulfur hexafuoride in a steel vessel having a capacity of 1250 ml at 70 c

Answer 1

volume of sulfur hexafluoride, SF6. = 1250ml = 1.25 L

temperature = 70°C = 273 + 70 = 343K

number of mole of sulfur hexafluoride = 0.25

use gas equation , PV = nRT

where P is the pressure, V is volume, n is number of mole , R is universal gas constant and T is temperature.

now, P = nRT/V

= 0.25 × 0.0821 × 343/1.25

= 0.0821 × 343/5

= 5.63 atm

Answer 2

Hey dear,

● Answer -
P = 5.7×10^5 Pa

● Explaination -
# Given -
V = 1250 ml = 1.25×10^-3 m^3
n = 0.25 mol
T = 70 °C = 343 K

# Solution-
For ideal gas equation,
PV = nRT
P = nRT / V
P = 0.25 × 8.314 × 343 / (1.25×10^-3)
P = 570340 Pa
P = 5.63 atm

Therefore, pressure exerted by sulphur hexafluoride on wall of container is 5.7×10^5 Pa.

Hope it is helpful...