Question: 0.37 litre of an ideal monatomic gas (Cv,m = 3R/2) initially at 51 °C and 36 atm pressure undergo an expansion against a constant external pressure of 1.06 atm, and do 2.3 kJ of work. The final pressure of the gas is 1.06 atm. Calculate the change in enthalpy, ΔH.
First, I calculated Vf using the equation W = -Pext(Vf-Vi) and found Vf to be 2170.18L.
Then, I used P1V1 = nRT1 to find n, which found to be 0.50075 moles.
Next, I used Tf = (PfVf)/(nR), and Tf = 55981.3 K.
Finally, I used ΔH = n(5R/2)(Tf - Ti) and found ΔH = 57929.19.
The answer I got is wrong :( can you please point out where I did wrong? Thank you so much
To find the change in enthalpy, ΔH, you need to use the equation:
ΔH = q + W
where q is the heat transferred and W is the work done.
Given that the work done is 2.3 kJ, you already have the value for W.
To find q, you can use the equation:
q = nCvΔT
where n is the number of moles, Cv is the molar heat capacity at constant volume, and ΔT is the change in temperature.
You've already calculated n to be 0.50075 moles.
The change in temperature, ΔT, can be found by converting the initial and final temperatures from Celsius to Kelvin:
ΔT = Tf - Ti
Initial temperature, Ti = 51 °C = 51 + 273.15 = 324.15 K
Final temperature, Tf = 55981.3 K (as you have calculated correctly)
Now, let's calculate q:
q = (0.50075 mol) * (3R/2) * (55981.3 K - 324.15 K)
Using R = 8.314 J/(mol·K), plug in the values and calculate q.
Once you have q, you can find ΔH using the equation:
ΔH = q + W
Substitute the values of q and W that you have calculated to find the correct value of ΔH.
To calculate the change in enthalpy (ΔH), you need to use the correct formula. In this case, the formula is ΔH = nCv,mΔT, where n is the number of moles of the gas, Cv,m is the molar heat capacity at constant volume, and ΔT is the change in temperature.
In your calculation, you correctly calculated the number of moles (n) to be 0.50075 moles.
However, you made a mistake in calculating the final temperature (Tf). The correct formula to use is PfVf = nRTf, where Pf is the final pressure, Vf is the final volume, n is the number of moles, R is the ideal gas constant, and Tf is the final temperature. Rearranging the formula, we get Tf = (PfVf)/(nR).
You correctly calculated Vf to be 2170.18 L, but you used the wrong equation to calculate Tf. Let's recalculate Tf using the correct formula:
Tf = (PfVf)/(nR)
= (1.06 atm * 2170.18 L) / (0.50075 mol * 0.0821 L·atm/(mol·K))
= 559.65 K
Now that we have the correct final temperature (Tf), we can calculate the change in enthalpy (ΔH):
ΔH = nCv,mΔT
= (0.50075 mol) * (3R/2) * (559.65 K - 51 °C + 273.15 K)
= (0.50075 mol) * (3R/2) * (862.80 K)
≈ 1454.92 J
Note that the conversion factor from Celsius to Kelvin is 273.15.
Therefore, the change in enthalpy (ΔH) is approximately 1454.92 J.