Respuesta :
Answer:
(A) Work done will be 87.992 KJ
(B) Work done will be 167.4 KJ
Explanation:
We have given mass of methane m = 4.5 gram = 0.0045 kg
Volume occupies [tex]V_1=12.7dm^3=12.7liters[/tex]
And volume is increased by [tex]3.3dm^3[/tex] so [tex]V_2=12.7+3.3=16liters[/tex]
Temperature T = 310 K
Pressure is given as 200 Torr = 26664.5 Pa
(a) At constant pressure work done is given by
[tex]W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj[/tex]
(b) At reversible process work done is given by [tex]W=nRTln\frac{V_2}{V_1}[/tex]
We have given mass = 4.5 gram
Molar mass of methane = 16
So number of moles [tex]n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125[/tex]
So work done [tex]W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J[/tex]
- The work done when the gas expands isothermally against a constant external pressure of 200 Torr is 87992.85J
- The work that would be done if the same expansion occurred reversibly is 167.55J
How to calculate work done?
According to this question, the following information is given:
- mass of methane = 4.5g = 0.0045kg
- volume (V1) = 12.7dm³
- Increase in volume (V2) = 12.7 + 3.3 = 16dm³
- T = 310K
- P = 200 Torr = 26664.5Pa
W = P (V2 - V1)
W= 26664.5 (16 - 12.7)
W = 87992.85J
However, at a reversible process, the work done is calculated as:
W = nRTlnV2/V1
Where;
- n = no of moles of methane = 4.5/16 = 0.2813mol
- R = gas law constant (8.314)
W = 0.2813 × 8.314 × 310 × ln (16/12.7)
W = 2.34 × 0.23 × 310
W = 167.55J
Therefore;
- The work done when the gas expands isothermally against a constant external pressure of 200 Torr is 87992.85J
- The work that would be done if the same expansion occurred reversibly is 167.55J.
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