If 6.81 mol of an ideal gas has a pressure of 2.99 atm and a volume of 94.35 L, what is the temperature of the sample?

504.57 K.

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 6.81 moles

Pressure (P) = 2.99 atm

Volume (V) = 94.35 L

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

Temperature (T) =.?

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

PV = nRT

2.99 × 94.35 = 6.81 × 0.0821 × T

282.1065 = 0.559101 × T

Divide both side by 0.559101

T = 282.1065 / 0.559101

T = 504.57 K.

Thus, the temperature of the ideal gas is 504.57 K.

your answer would be “dry”

the formula for the conversion of temperature from degree celsius to kelvin is:

the sample of water at is converted to kelvin temperature by using the above formula as:

on comparing the two samples of water at or we can say that at and at respectively. the difference between the temperatures of two samples of water is:

hence, one can say that the sample of water at is greater than the sample of water at by .