Question

In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60×105Pa3.60 \times 10^{5} \mathrm{Pa} and 300 K, expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a pV-diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

Solution

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Given

We are given an ideal monatomic gas with a number of moles nn = 1.20 mol at initial pressure p1=2.60×105Pap_{1} = 2.60 \times 10^{5} \,\text{Pa} and initial temperature T1T_{1} = 300 K. The gas expands until its volume triples which means V2=3V1V_{2} = 3 V_{1}. We want to calculate the work done of the process if the expansion is (a) isothermal, (b) adiabatic, and (c) isobaric.

Solution

(a) Isothermal process means the temperature is constant during the process where ΔT\Delta T = 0 while the pressure and the volume are changed. So the work done for the isothermal process is given by

W=V1V2pdVW = \int_{V_{1}}^{V_{2}} p dV

Where pp could be extracted from ideal gas law by (p=nRTV)\left(p=\dfrac{nRT}{V}\right). Now plug this value of pp into equation (1) and take the constant term nRTnRT outside the integration and complete the integration where equation (1) will be

W=nRTV1V21VdV(1V)=lnVW=nRTln(V2V1)(1*)\begin{gathered} W = nRT \int_{V_{1}}^{V_{2}} \dfrac{1}{V} dV \text{$\int \left(\dfrac{1}{V}\right) = \ln V$}\\ W = nRT \ln \left(\dfrac{V_{2}}{V_{1}} \right) \tag{1*} \end{gathered}

Now we can plug our values for n,R,Tn, R, T and V2V_{2} into equation (1*) to get the work done during the isothermal process where RR is the gas constant and equals 8.314 J/molK8.314 \mathrm{~J/mol\cdot K} (See Appendix F) and V2=3V1V_{2} = 3V_{1}

W=nRTln(V2V1)=1.20mol×8.314 J/molK×300Kln(3V1V1)=3288J\begin{aligned} W &= nRT \ln \left(\dfrac{V_{2}}{V_{1}} \right) \\ &= 1.20 \,\text{mol} \times 8.314 \mathrm{~J/mol\cdot K} \times 300 \,\text{K} \ln \left(\dfrac{3 \cancel V_{1}}{\cancel V_{1}} \right)\\ & = \boxed{3288 \,\text{J}} \end{aligned}

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