Question

A sample of helium gas has a volume of 0.180L, a pressure of 0.800a and a temperature of 29 C. What is the new temp or gas of volume of 90mL and pressure of 3.20 atm

Answer 1

Answer:

The new temperature of the gas is 604K

Explanation:

Assuming standard temperature and pressure, to calculate the temperature of the gas we use the general gas equation;

Step 1 :write the general gas equation

P1V1/ T1 = P2V2/T2

Step 2: Write out the values, covert the necessary values to the standard values.

P1 = 0.800atm.

V1 = 0.180L

T1 = 29°C = 273 + 29 = 302K

P2 , 3.20atm

V2 = 90mL = 90 * 10^-3L = 0.09L

Step 3: Solve for T2

The new temperature T2 of the gas is:

T2 = P2V2T1/ P1V1

T2 = 3.20 * 0.09 * 302 / 0.800 * 0.180

T2 = 86.976 / 0.144

T2 = 604K

The new temperature is of the gas is 604K.