A sample of ideal gas is expanded to twice its original volume of 1.00 m3 in a quasi-static process for which P = áV2, with á = 2.50 atm/m6, as shown in the figure. How much work is done by the expanding gas?
i tried using the eq:
w=-1/3(2.50atm/m^6)(1.013x10^5)[(2.00m^3)-(1.00m^3)]
i end up getting 5.84x10^6 but this is not the right answer help?
Is V2 supposed to represent V^2?
The work done by the gas equals the integral of P dV where V goes from V1 to V2 = 2V1. They tell you P as a function of V.
Just do the integration of P(V) dV. To get the answer in Joules, you will have to use P units of Newton/m^2, not atm.
W = Integral of a V^2 dV
V1 = 1.00 m^3 to V2 = 2.00 m^3
= (a/3) (V2^3 - V1^3) = (a/3)(8 - 1)m^9
where
a = 2.5*1.01*10^5 N/m^8
To solve this problem, you need to calculate the integral of the PdV equation over the given volume change. The equation you mentioned, w = -1/3αP[(Vf^3) - (Vi^3)], is correct, but it seems that there might be a mistake in the calculations.
Let's break down the steps to calculate the work done by the expanding gas:
1. Given:
Initial volume, Vi = 1.00 m^3
Final volume, Vf = 2.00 m^3
α = 2.50 atm/m^6 (constant)
Pressure, P = αV^2
2. Calculate the work done, w:
w = ∫ PdV (from Vi to Vf)
= ∫ αV^2dV (from Vi to Vf)
Since P = αV^2, we can rewrite the equation:
w = ∫ αV^2dV
= α∫ V^2dV (from Vi to Vf)
3. Solve the integral:
w = α∫ V^2dV
= α * [(1/3)V^3] (from Vi to Vf)
= α * [ (1/3)(Vf^3 - Vi^3) ]
Plug in the given values:
w = 2.50 atm/m^6 * [ (1/3)(2.00^3 - 1.00^3) ]
= 2.50 atm/m^6 * (1/3) * (8 - 1)
= 2.50 * 2.00 atm * m^3
= 5.00 atm * m^3
Lastly, convert the units to Joules:
1 atm * m^3 = 101.325 J
5.00 atm * m^3 = 5.00 * 101.325 J
≈ 506.625 J
Therefore, the correct value for the work done by the expanding gas is approximately 506.625 Joules.
To determine the work done by the expanding gas in this process, you correctly used the formula:
w = ∫P dV
However, there seems to be a slight error in the calculations.
Given:
Initial volume, V1 = 1.00 m^3
Final volume, V2 = 2.00 m^3
Pressure function, P = αV^2, where α = 2.50 atm/m^6
To evaluate the integral, we need to express pressure (P) in terms of volume (V). Using the given pressure function:
P = αV^2
Since α is constant, we can rewrite α as:
α = P/V^2
Substituting this back into the formula for work:
w = ∫P dV
w = ∫(P/V^2) dV
Now we can integrate this expression over the range of volumes from V1 to V2:
w = ∫(P/V^2) dV (from V1 to V2)
Evaluating this integral:
w = ∫(P/V^2) dV
w = ∫(2.50 atm/m^6)(1/V^2) dV (from V1 to V2)
Integrating (1/V^2) with respect to V gives (-1/V):
w = (-2.50 atm/m^6) ∫(1/V) dV (from V1 to V2)
w = (-2.50 atm/m^6) [ln(V)] (from V1 to V2)
w = (-2.50 atm/m^6) [ln(V2) - ln(V1)]
Now, substitute the given values:
w = (-2.50 atm/m^6) [ln(2.00) - ln(1.00)]
w = (-2.50 atm/m^6) [ln(2.00/1.00)]
w = (-2.50 atm/m^6) [ln(2.00)]
w = (-2.50 atm/m^6) [0.693]
w = -1.73 atm*m^6
Note: The negative sign indicates work done by the gas on the surroundings.
Now, converting the units from atm*m^6 to J (Joules):
1 atm * m^3 = 101.325 J
w = -1.73 atm*m^6 * 101.325 J/(1 atm * m^3)
w = -1.73 * 101.325 J*m^3
w = -175.27 J
Therefore, the work done by the expanding gas is approximately -175.27 Joules.