Question

At what temperature in K will 0.750 moles of oxygen gas occupy 10.0 L and exert 2.50 atm of pressure

Answer 1

Answer: 406 K

Step-by-step explanation:

We can rewrite the ideal gas law to solve for T:

PV = nRT

T=PV / nR

We are given the following from the problem:

n=0.750 mol P=2.50 atm V=10.0 L

Plugging in our values and using R=0.08206 L⋅atm / K⋅mol we get:

T=(2.50 atm)(10.0 L) / (0.750 mole)(0.08206L ⋅ atm ⋅ mole K) = 406 K

Answer 2

406 K.

Step-by-step explanation:

The following data were obtained from the question:

Number of mole (n) = 0.750 mole

Volume (V) = 10.0 L

Pressure (P) = 2.50 atm

Temperature (T) =.?

Note: Gas constant (R) = 0.0821 atm.L/Kmol

The temperature, T can be obtained by using the ideal gas equation as follow:

PV = nRT

2.5 x 10 = 0.75 x 0.0821 x T

Divide both side by 0.75 x 0.0821

T = (2.5 x 10) /(0.75 x 0.0821 )

T = 406 K.

Therefore, the temperature is 406 K.