Temperature of 1.4 moles of gas's with pressure of 3.25 atmosphere in a 4.738-liter container

-139 °C

Step-by-step explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT Divide each side by nR

T = (pV) / (nR)

dа ta:

p = 3.25 atm

V = 4.738 L

n = 1.4 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculation:

T = (3.25 * 4.738) / (1.4 * 0.082 06)

T = 134 K

T = (134 - 273.15) °C

T = - 139 °C