91影视

ideal monatomic gas has a number ofmoles in= 1.20 mol

initial pressure p1=2.60脳105Paand initial temperature T1= 300 K. The gas expands until its volume triples which means V2=3V1.

a)

the temperature is constant during Isothermal process T螖T = 0 while the pressure and the volume are change. So, the work done is,

$\mathrm{W}={鈭�}_{{\mathrm{V}}_1}^{{\mathrm{V}}_2}\mathrm{pdV}$鈥︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹��. (1)

As,$\mathrm{p}=\frac{\mathrm{nRT}}{\mathrm{V}}$

$\mathrm{W}=\mathrm{nRT}{鈭�}_{{\mathrm{V}}_1}^{{\mathrm{V}}_2}\frac{1}{\mathrm{V}}d\mathrm{V}$

$\mathrm{W}=\mathrm{nRTIn}\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)$鈥︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�� (2)

put values for n, R, Tn,R,T in equation (2)

$$\begin{gathered} \mathrm{W}=\mathrm{nRTIn}\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right) \\ \mathrm{W}=1.20\mathrm{mol}脳8.314\mathrm{J}/\mathrm{mol}脳\mathrm{K}脳300\mathrm{In}\left(\frac{3{\mathrm{V}}_2}{{\mathrm{V}}_1}\right) \\ \mathrm{W}=3288\mathrm{J} \end{gathered}$$

Thus,the work done by the gas if isothermal W=3288J

b)

InAdiabatic process there is no heat transferQ=0. The work done is,

$\mathrm{W}=\mathrm{Q}鈥�鈭�\mathrm{U}$

Here,

$\mathrm{W}=鈥�鈭�\mathrm{U}$

$鈭�\mathrm{U}=鈥�{\mathrm{nC}}_{\mathrm{V}}鈭�\mathrm{T}$鈥︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹��. (3)

Take$鈭�\mathrm{T}={\mathrm{T}}_2鈥�{\mathrm{T}}_1$ and${\mathrm{C}}_{\mathrm{V}}=3\mathrm{R}/2$

Thus,

${\mathrm{T}}_2{\left({\mathrm{V}}_2\right)}^{\mathrm{纬}-1}={\mathrm{T}}_1{\left({\mathrm{V}}_1\right)}^{\mathrm{纬}-1}$鈥︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�� (4)

y is the ratio of molar heat capacities,

$\mathrm{纬}={\mathrm{C}}_{\mathrm{p}}/{\mathrm{C}}_{\mathrm{V}}=\frac{5\mathrm{R}/2}{3\mathrm{R}/2}=1.67$

put these values it in equation (4),

$$\begin{gathered} T_2=T_1\left(\frac{V_1}{V_2}\right) \\ {T}_2=300{\left(\frac{V_1}{3V_1}\right)}^{1.67-1} \\ {T}_2=144K \end{gathered}$$

Put all the obtained values in equation (3)

$$\begin{gathered} \mathrm{W}=鈥�\mathrm{nC}鈭�\mathrm{T} \\ \mathrm{W}=鈥�\mathrm{n}\left(\frac{3\mathrm{R}}{2}\right)\left({\mathrm{T}}_2鈥�{\mathrm{T}}_1\right) \\ \mathrm{W}=鈥�1.20\mathrm{mol}\left(\frac{3脳8.314\mathrm{J}/\mathrm{mol}}{2}\right)\left(144\mathrm{K}-300\mathrm{K}\right) \\ \mathrm{W}=2335\mathrm{J} \end{gathered}$$

Thus, the work done by the gas if adiabatic is W=2355J

(c)

the pressure during the process is constant in Isobaric process, and the work done is,

$$\begin{gathered} W=p鈭�V \\ W=p\left(V2-V1\right) \end{gathered}$$

$W=p\left(2V1\right)$鈥︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�� (5)

We are not given the value of V1 so use the ideal gas law as pV1=nRT1 and equation (5) will be

$\mathrm{W}=2\mathrm{nRT}1$鈥︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�︹�� (6)

put values for n, Rn,R and }T1 into equation (6)

$$\begin{gathered} \mathrm{W}=2\mathrm{nRT}1 \\ \mathrm{W}=2脳1.20\mathrm{mol}脳8.314\mathrm{J}/\mathrm{mol}脳\mathrm{K}脳300\mathrm{K} \\ \mathrm{W}=5986\mathrm{J} \end{gathered}$$

Thus, the work done by the gas is isobaric. Is W=6000J

( d)

here the greatest work done is during the isobaric process also the pressure is constant and the least one is during the adiabatic process (b) when the heat transfer is zero. Less work done shows more curvature.

Thus,the absolute value of the work done by the gas is greatest.

(e) by first law of thermodynamics heat transfer will be for processes with great work. Also, the heat transfer will be for isobaric process c . Heat transfer will be for adiabatic process b.

Thus, the absolute value of the heat transfer is greatest

(f) The change in internal energy 螖U depends on the change in temperature

$$\begin{gathered} T_2=T_1\left(\frac{V_2}{V_1}\right) \\ {T}_2=300\left(\frac{3V_1}{V_1}\right) \\ {T}_2=900K \end{gathered}$$

螖T = 600 K. the temperature is constant during the process, so the change in temperature 螖T=0and therefore螖U=0and show the least change in internal energy and the greatestfor isobaric.

Thus, the absolute value of the change in internal energy of the gas is least