A 1.5 L pocket of air with a temperature of 295 K rises in the atmosphere. What will be the volume of the air pocket of the t temperature decreased to 2 celsius and the pressure is not changed.

1.4 L.

Explanation:

Applying Charles law,

V/T = V'/T' Equation 1

Where V = Initial volume, V' = Final volume, T = Initial Temperature in Kelvin, T' = Final Temperature in Kelvin.

Make V' the subject of the equation

V' = (V/T)T' Euqation 2

Given: V = 1.5 L, T = 295 K, T' = 2 °C = (2+273) K = 275 K

Substitute these values into equation 2

V' = (1.5/295)275

V' = 1.398

V' ≈ 1.4 L

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

the answer is 0.8m.

by: nicholas