Answer:
- 1384.7 J.
Explanation:
STEP ONE: make use of the equation of state to determine the amount of pressure.
The equation of state is given below as;
NRT = PV. Where P = pressure, V = volume, R = gas constant, T= temperature and N = number of moles.
Therefore, making P the subject of the formula, we have;
P = NRT/V.
Thus, P = (1 + a/V + b/V^2) [(RT)/V].
Note that a = -265 cm^3/g = - 0.265 m^3/kg and b = 30250 cm^6/g = 0.030250 m^6/kg.
Hence, P = ( 1 - 265/V + 0.030250/V^2) {(RT)/V}.
P = ( 1/V - 265/V^2 + 0.030250/V^3) × RT.
STEP TWO: Calculate for the volume using the equation of state in step one above.
V = ( 1 - 265/V+ 0.030250/V^2) × nRT/P
V = ( 1 - 265/V+ 0.030250/V^2) × (1 × 0.08205 × 460/50) = 0.39L.
Thus, we have that the V above = V1 = 0.39L.
Also, V = ( 1 - 265/V+ 0.030250/V^2) × (1 × 0.08205 × 460/100) = 0.1769L.
Workdone, W = Integration of - PdV.
Taking V1 and V2 as the lower and upper limits.
W = - ∫ PdV = - ∫ ( 1/V - 265/V^2 + 0.030250/V^3) dV.
W = - 37.7 + (-0.82237 + 0.8264 - 0.3862). = 13.67 × 101325 × 10^-3 = - 1384.7J.
In this exercise we have to use integration to calculate the work of the molecule, so:
[tex]1384.7 J.[/tex]
Make use of the equation of state to determine the amount of pressure. The equation of state is given below as:
[tex]PV=NRT[/tex]
Where:
Therefore, making P the subject of the formula, we have;
[tex]P = \frac{NRT}{V}[/tex]
Thus,
[tex]P = (1 + a/V + b/V^2) [(RT)/V][/tex]
Note that:
[tex]a = -265 \ cm^3/g = - 0.265\ m^3/kg \\ b = 30250\ cm^6/g = 0.030250 \ m^6/kg.[/tex]
Hence,
[tex]P = ( 1 - 265/V + 0.030250/V^2) {(RT)/V}.\\P = ( 1/V - 265/V^2 + 0.030250/V^3) *( RT)[/tex]
Calculate for the volume using the equation of state in step one above:
[tex]V = ( 1 - 265/V+ 0.030250/V^2) ( nRT/P)\\V = ( 1 - 265/V+ 0.030250/V^2)* ((1)( 0.08205 )( 460/50)) = 0.39L[/tex]
Thus, we have that the V above : [tex]V_1 = 0.39L[/tex]
Also,
[tex]V = ( 1 - 265/V+ 0.030250/V^2) * ((1)*( 0.08205)*( 460/100)) = 0.1769L[/tex]
Work done,
[tex]W = \int\limits^a_b {-P} \, dV[/tex]
Taking [tex]V_1[/tex] and [tex]V_2[/tex] as the lower and upper limits:
[tex]W = - \int\limits {P} \, dV = - \int\limits { 1/V - 265/V^2 + 0.030250/V^3} \, dV\\W = - 37.7 + (-0.82237 + 0.8264 - 0.3862). = (13.67)*( 101325)*( 10^-3) = - 1384.7J[/tex]
See more about work at brainly.com/question/1374468
