5. A sample of water vapor occupies a volume of 10.5 L at 200C and 100.0 kPa.
What volume will the water vapor occupy when it is cooled to 27C if the
pressure remains constant?
6. What is the volume occupied by 0.355 mole of nitrogen gas at STP?
7. What is the volume of a container that holds 25.0 g of carbon dioxide gas
at STP?
Dr W
Favorite Answer
*** 5 ***
I always recommend these steps for a 2 state gas law problem like you have here
1) start with the equation P1V1 / (n1T1) = P2V2 / (n2T2)
2) identify and cancel anything constant
3) rearrange and solve for you unknown
V and T vary
P and moles (n) remain constant
so...
P1V1 / (n1T1) = P2V2 / (n2T2)
becomes...
V1 / T1 = V2 / T2
rearranging and solving...
V2 = V1 x (T2 / T1) = 10.5L x (300K / 473K) = 6.66L
*** 6 ***
PV = nRT
V/n = RT/P
@STP, T=273K and P=1atm...making
V/n = (0.08206 Latm/moleK) x (273.15K) / (1atm) = 22.4 L/mole... recognize that number?
so...
0.355 moles x (22.4 L/mole) = 7.95L
alternately
PV = nRT
V = nRT/P = (0.355moles) x (0.08206 Latm/moleK) x (273.15K) / (1atm) = 7.95L
your choice
*** 3 ***
25.0g CO2 x (1 mole CO2 / 44.01g CO2) x (22.4 L CO2 / mole CO2@STP) = 12.7 L CO2
joh
even nonetheless the entropy decreases, the enthalpy is detrimental adequate to make the reaction spontaneous. The device is DeltaG=DeltaH-(T)DeltaS G is the loose potential T is the temperature S is the entropy if DeltaG is poor, then the reaction is spontaneos. because of the undeniable fact that DeltaS is undesirable, the -TDeltaS term is advantageous this confident term is attempting to make the reaction nonspontaneous if the reaction is nicely spontaneous, then the DeltaH(enthalpy) must be low adequate to offset the advantageous TDeltaS. subsequently the enthalpy is in basic terms very poor