How Is Heat Absorbed by Gas in a Pressure and Volume Increase Process?

[sun]

Heat absorbance: PLS PLS HELP ASAP!

Homework Statement

A quantity of a monatomic ideal gas undergoes a process in which its pressure is increased by a factor of n = 3 and its volume is increased by a factor of m = 6 as shown in Figure P12.12. What is the heat absorbed by the gas during this process? (Hint: See Problem 1. Answer in terms of P0V0.)

http://www.webassign.net/sercp/p12-12alt.gif

this is what I've done so far, but i keep getting it incorrect.
work (W), which equals -1/2 (n*Po-Po)(m*Vo-Vo) - Po(Vo-Vo)
delta U = (3/2) [(m*Po)(n*Vo) - PoVo]

delta U=(3/2)[(6Po)(3Vo)-PoVo]=(3/2)(17PoVo)
W=(-1/2)((3Po-Po)(6Vo-Vo)-Po(Vo-Vo))=(-1/2)((2Po)(5Vo)-(Po(Vo-Vo))=-5PoVo

U=Q+W

I solved for W and got 30.5xPoVo.

I'm clearly doing some incorrect algebra somewhere. Could i please get some help?

thanks

 

[dynamicsolo]

this is what I've done so far, but i keep getting it incorrect.
work (W), which equals -1/2 (n*Po-Po)(m*Vo-Vo) - Po(m*Vo-Vo)

Since the process path is a straight line, the work on the gas is just the negative of the area of the trapezoid under the line, which is what you have. (I corrected your second term.) If you put in the values for m and n, you get -10PoVo.

delta U = (3/2) [(m*Po)(n*Vo) - PoVo]

We'll call the number of moles C for the moment (it cancels out). So To = PoVo/CR and T' = 18PoVo/CR , so

delta U = (3/2)·C·R· (17PoVo/CR) = (51/2)PoVo = 25.5PoVo

deltaU=Q+W

so Q = deltaU - W = 25.5PoVo - (-10PoVo) = 35.5PoVo .

It's that second term in your work equation: you dropped the factor n in the volume difference.