Question

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 atm. 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.

Answer 1

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$=P\Delta V$

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

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

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

$\Delta q=\Delta W=9.852 J$

$\Delta U=0$