Solve Work Problem: 2.03 mol He, 295 K, 0.350 atm to 1.00 atm

[nemzy]

here is the problem:

A 2.03 mol sample of helium gas initially at 295 K and 0.350 atm is compressed isothermally to 1.00 atm. Assume that the helium behaves as an ideal gas.


what is the work done on the gas.




well i know that w=-p*change in volume

i can figure out the final and initial volume using the ideal gas law pv=nrt at the initial and final pressures...so i know what the change in volumeis, but what would the pressure be when calculating the work? i used both 1 atm and .35 atm and i stil get the wrong answer...here is my work

final volume: pv=nrt
p=1.8 atm
n=2.18 mol
r= .0821 liters*atm/mol*k
t=305 k

final volume = .0303 m^3

initial volume: pv=nrt
p=.505 atm
n=2.18
r=.0821 liters*atm/mol*k
t=305 k

initial volume = .108 m^3

change in volume = .0778 m^3

work = -p(change in volume)


the answer is suppose to be 7.03 kj, which means that if my volume calculations were right, the pressure is suppose to be .893 atm?? does anyone know where i went wrong?

 

[wisky40]

Work is defined as integral of pdv => w=Int{pdv}=Int{nRT(dv/v)},where t is constant bacause the process is isothermic => w=(nRT)In{(v_f)/(v_i)} and
v_f/v_i = p_i/p_f.
There are some other problems like"A 2.03 mol sample..." and then you used n=2.18 also "... at 295 K..." T= 305k... and so on

 

[ikhan]

incorrect on the W there...W=(nrt)ln(v_i/v_f)...