Mere approximation problem or something else? Thermodynamics

[mooncrater]

Homework Statement

The question is :
Calculate the work done in joules when 1.0 mole of N₂H₄ decomposed against a pressure of 1.0 atm at 300 K for the equation:
3N₂H₄
(l)→4NH₃(g)+2N₂(g)

Homework Equations

None

The Attempt at a Solution

I did it like as:
Assuming 100% dissociation of N₂H₄
4/3 moles of NH₃ and 2/3 moles of N₂
will be formed. That means a total of 2 moles of gas will be formed , which will have a volume of 2×22.4L at STP.
(SINCE AT STP VOLUME OF 1 MOLE OF GAS=22.4L)
Which will be equal to 44.8 ×10^-3 m³
So work done =PΔV=10⁵×44.8×10^-3=4480J (with a negative sign)
But its answer is -4988.4 J which is much different from mine. What I think is that mere approximations in the values of Molar volume at STP and pressure change from atm to N/m² cannot cause such a large deviation. So is my method incorrect? Or the given answer?

 

[Quantum Defect]

Homework Statement

The question is :
Calculate the work done in joules when 1.0 mole of N₂H₄ decomposed against a pressure of 1.0 atm at 300 K for the equation:
3N₂H₄
(l)→4NH₃(g)+2N₂(g)

Homework Equations

None

The Attempt at a Solution

I did it like as:
Assuming 100% dissociation of N₂H₄
4/3 moles of NH₃ and 2/3 moles of N₂
will be formed. That means a total of 2 moles of gas will be formed , which will have a volume of 2×22.4L at STP.
(SINCE AT STP VOLUME OF 1 MOLE OF GAS=22.4L)
Which will be equal to 44.8 ×10^-3 m³
So work done =PΔV=10⁵×44.8×10^-3=4480J (with a negative sign)
But its answer is -4988.4 J which is much different from mine. What I think is that mere approximations in the values of Molar volume at STP and pressure change from atm to N/m² cannot cause such a large deviation. So is my method incorrect? Or the given answer?

What temperature is "STP" -- how does that compare with the 300 K of the original problem?

 

[mooncrater]

Okay...that was a silly mistake...STP is for 273K not 300K . Now I will never forget it. Thank you!

 

[Chestermiller]

It seems to me there is something wrong here. If the initial number of moles were 1, and the final number of moles were 2, then at 273K, the change in volume would have been 22.4 liters, not 44.8 liters. So it seems to me that the "correct answer" is a factor of 2 too high. What do you guys think?

Chet

 

[mooncrater]

But $N_2H_4$ is a liquid at initial conditions.

 

[Chestermiller]

But $N_2H_4$ is a liquid at initial conditions.

Ah. Thanks. That explains it. Duh!

Chet

 

[Raghav Gupta]

Ah. Thanks. That explains it. Duh!

Chet

Sir I think you knew it but asked the question to check mooncrater.

 

[Chestermiller]

Sir I think you knew it but asked the question to check mooncrater.

You're very kind, but, no, I didn't notice that (l) in the problem statement. And I definitely didn't know from prior experience that N2H4 is a liquid at ordinary conditions.

Chet