Explanation:
The balanced chemical equation of the reaction is:
From the balanced chemical equation,
1 mole of propane forms ------ 3 mol. of gas.
The molar mass of propane is 44.1 g/mol.
One mole of any gas at STP occupies --- 22.4 L.
Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.
Answer:
Thus, 67.2 L of CO2 is formed at STP.
Answer: Volume of the gas at STP is 22.53 L.
Explanation:
Given : Volume = 125 mL (as 1 mL = 0.001 L) = 0.125 L
Temperature =
Pressure =
According to the ideal gas equation, the volume of given nitrogen gas is calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
Hence, volume of the gas at STP is 22.53 L.
Explanation:
As, for reversible isothermal expansion the formula is as follows.
W =
Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.
As the given data is as follows.
R = 8.314 J/(K mol), T = 298 K ,
= 8.34 L,
= 2.67 L
Now, putting the given values into the above formula as follows.
W =
=
=
= -2818.68 J
Hence, work for path A is -2818.68 J.
Step 1: When there is no change in volume then W = 0
Hence, for step 1, W = 0
Step 2: As, the gas is allowed to expand against constant external pressure = 1.00 atm.
So, W =
Now, putting the given values into the above formula as follows.
W =
=
= -5.67 atm L
As we known that, 1 atm L = 101.33 J
Hence, work will be calculated as follows.
W =
= -574.54 J
Therefore, total work done by path B = 0 + (-574.54 J)
W = -574.54 J
Hence, work for path B is -574.54 J.