Pressure and volume of a gas

[turophile]

Here's the problem:

Assume that the pressure and volume of a gas are related by pV^(1.4) = C, where C is a constant. Find the work done in compressing 1024 cu. ft. of gas at 18 lbs./sq. in. to 243 cu. ft. What is the final pressure?

Here's my work:

V[1] = 1024
p[1] = 18
V[2] = 243

p = f(V) = C * V ^ (-1.4)
C = p[1] * V[1] = 18 * 1024 ^ (1.4)

W = C * integral{243 to 1024} V ^ (-1.4) dV
= [(5 * C) / 2] * [243 ^ (-0.4) - 1024 ^ (-0.4)]
= (5/2) * [18 * 1024 ^ (1.4)] * [243 ^ (-0.4) - 1024 ^ (-0.4)]
= [(5 * 18) / 2] * 1024 * [1024 ^ (0.4) * 243 ^ (-0.4) - 1]
= 45 * 1024 * [(1024/243)^0.4 - 1)]
= 46080 * (16/9 - 1)
= 46080 * (7/9)
= 5120 * 7
= 35840 (in units of 144 ft.-lbs.)
= 35840 * 144 = 5160960 ft.-lbs.

p[2] = C * (243) ^ (-1.4)
= 18 * 1024 ^ (1.4) * 243 ^ (-1.4)
= 18 * (1024/243) ^ (1.4)
= 18 * (4/3) ^ 7
= (2 * 4 ^ 7)/(3 ^ 5)
= (2 * 16384)/243
= 32768/243
= 134 + 206/243 lbs./sq. in.

My question:

I checked my answers with those given in my textbook, and my calculation for W matches but not my calculation for p[2]. The textbook gives p[2] = 75 + 23/27 lbs./sq. in., which is 18 * 1024 ^ (1) * 243 ^ (-1), not 18 * 1024 ^ (1.4) * 243 ^ (-1.4), the answer I calculated. So my textbook could be wrong, or I could be. If it's me, where did I go wrong?

 

[galactus]

Unless we're both incorrect, I get the same as you.

Using:

\(\displaystyle \text{Work}=\frac{C}{.4}\left(\frac{1}{V_{1}^{.4}}-\frac{1}{V_{2}^{.4}}\right)\)

Using this formula gives the same answer as per your first solution: 5160960

Now, using \(\displaystyle p_{2}\cdot V_{2}^{1.4}=C\) and solving for p2,

I get the same result as you do.

 

[turophile]

Thanks!