Question

2. What would the temperature of 0.734 moles of chlorine gas be at 1.83 atmospheres in a 10.2 liter container? (R = 0.0821 L·atm/mol·K)

Answer 1

p = 1.83 atm

n = 0.734 moles

V = 10.2 L

R = 0.0821 L·atm/mol·K

Ideal gas law:

pV = nRT

$T = \frac{pV}{nR} \\ T = \frac{1.83 \times 10.2}{0.734 \times 0.0821} \\ T = 309.75 \;K \\ T = 309.75-273.15 = 36.6 \;ºC \\ Answer: 36.6\; ºC$