if i have 2.4 moles of gas held at a temperature of 97 c and in a container with a volume of 45 L , what is the pressure of the gas

1.62 L

Explanation:

T= 97+273.15= 370.15

R= 0.08206 atm/mol⋅K

V= 45 L

n= 2.4 mol

P= (n⋅R⋅T)/V

= (2.4 x 0.08206 x 370.15)/(45) = 1.61997 = 1.62

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm