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rusak2 [61]
3 years ago
5

One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at

m. Determine the value of q, w.
(b) Now assume the gas expands by the same amount again isothermally but now irre-

versibly against 1 atm pressure (instead of reversible expansion) and calculate again q, w, delta U and delta H.
Physics
1 answer:
Zolol [24]3 years ago
8 0

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature(T_i)=300 K

P_i=10 atm

P_f=2 atm

Work done in iso-thermal process=P_iV_iln\frac{P_i}{P_f}

P_i=initial pressure

P_f=Final Pressure

W=10\times 2.463\times ln\frac{10}{2}=39.64 J

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is=P\Delta V

V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L

V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L

\Delta W=1\times (12.315-2.463)=9.852 J

\Delta q=\Delta W=9.852 J

\Delta U=0

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