- #1
FaraDazed
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Homework Statement
a ideal monoatomic gas initally has a temperature of 315K and a pressure of 6.87atm . It then expands from a volume of 440cm^3 to volume 1550cm^3 .
If the expansion is isothermal, what is (A) the final pressure (in atm's) and (B) the work done by he gas.
Homework Equations
[itex]PV^{\gamma}=[/itex] constant
The Attempt at a Solution
For the final pressure I just did the following, no idea if this is correct mind you.
[itex]
P_iV_i^{\gamma}=P_fV_f^{\gamma} \\
P_f=\frac{P_iV_i^{ \gamma } } { V_f^{ \gamma } }=\frac{ (6.87 \times 1.01 \times 10^5)(0.44)^{ \frac{7}{5} } }{1.55^{\frac{7}{5}}} = 119027.5 Pa \\
\frac{119027.5}{1.01 \times 10^5} = 1.178 atm \\
[/itex]
Then for the work done i did the below,
I first found the constant, c, the value of PV^7/5, which was 219842.6
then did
[itex]
W=\int_{V_i}^{V_f} P \,\, dV =\int_{V_i}^{V_f} \frac{c}{V^{\gamma}} = c \int_{V_i}^{V_f} V^{-\gamma}
[/itex]
I forgot to mention, our lecturer told us to assume gamma is 7/5 for this case.
[itex]
W =c \int_{V_i}^{V_f} V^{-\gamma} = [c(\frac{5V^\frac{12}{5}}{12} ) ]_{V_i}^{V_f} \\
W = [(219842.6)\frac{5(1.178 \times 1.01 \times 10^5)^{\frac{12}{5}}}{12} ] - [(219842.6)\frac{5(6.87 \times 1.01 \times 10^5)^{\frac{12}{5}}}{12} ] \\
W= (1.39 \times 10^{17}) - (9.57 \times 10^{18} ) = -9.43 \times 10^{18} J
[/itex]
Extremely new to this material and have not got my head around the material yet, so I would be suprised if this is correct. The value does seem rather low. What I am a little concerned with is the fact they explicitly state the temperature of the gas, but I have not used temperature in any of the equations/formulas used. The question does go on further to ask for the final volume and work if the process was adiabatic rather than isothermal, so it may be used there?
EDIT: Just realized a school boy error with the integration. Have gone back and already amended it as of now.
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