What is the approximate temperature of 1.4 moles of a gas with a pressure of 3.25 atmospheres in a 4.738-liter container

134K

Explanation:

Using the ideal gas law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas constant (0.0821 Latm/Kmol)

T = temperature (K)

Based on the information provided, n = 1.4moles, P = 3.25atm, V = 4.738L, T = ?

3.25 × 4.738 = 1.4 × 0.0821 × T

15.3985 = 0.11494T

T = 15.3985/0.11494

T = 133.969

Approximately;

T = 134K

b) mutualism

hope this : )

the answer is b. dilution