At what temperature will 2.40 moles of chlorine gas exert a pressure of 2.70 atm at a volume of 0.750 L?

—262.71°C

Explanation:

Step 1:

The following data were obtained from the question:

Number of mole (n) = 2.4 moles

Pressure (P) = 2.70 atm

Volume (V) = 0.750 L

Temperature (T) =?

Gas constant (R) = 0.082atm.L/Kmol

Step 2:

Determination of the temperature.

Using the ideal gas equation, the temperature can be obtained as follow:

PV = nRT

2.7 x 0.750 = 2.4 x 0.082 x T

Divide both side by 2.4 x 0.082

T = (2.7 x 0.750) /(2.4 x 0.082)

T = 10.29K

Step 3:

Conversion of Kelvin temperature to celsius temperature.

Temperature (celsius) = temperature (Kelvin) - 273

temperature (Kelvin) = 10.29K

Temperature (celsius) = 10.29 - 273

Temperature (celsius) = —262.71°C

10.28Kelvin

Explanation:

Using the ideal gas equation;

PV = nRT

P is the pressure

V is the volume of the gas

n is the number of moles

T is the temperature in Kelvin

R is the Gas constant

Given n = 2.4moles

P = 2.70atm

V = 0.750L

R = 0.0821atm.L/mol.K

From the formula above:

T = PV/nR

T = 2.70×0.750/2.4×0.0821

T = 2.025/0.197

T = 10.28K