Amass of gas has a volume of 4 m3, a temperature of 290 k, and an absolute pressure of 475 kpa. when the gas is allowed to expand to 6.5 m3, its new temperature is 277 k. what's the absolute pressure of the gas after expansion?

a. 104.1 kpa
b. 293.9 kpa
c. 279.2 kpa
d. 178.5 kpa

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At number of moles the value of PV/T is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1/T1 = P2V2/T2

P2 = P1 (V1) (T2) / (T1) (V2)

P2 = 475 kPa (4 m^3) (277 K) / (290 K) (6.5 m^3)

P2 = 279.20 kPa

Therefore, the changes in the temperature and the volume lead to a change in the pressure of the system which is from 475 kPa to 279.20 kPa. So, there is a decrease in the pressure.

the ratio of the conjugate base to the acid required to attain the desired ph.