Q#19.1
Two moles of an ideal gas are heated at constant pressure from T = 27$^0C$ to T = 107$^0C$.
(a) Draw a pV-diagram for this process.
(b) Calculate the work done by the gas.
Answer:
(a) The pressure is constant and the volume increases.
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| Fig.19.1 |
(b) W = $\int_{V_1}^{V_2}pdV$
Since p is constant,
W = p$\int_{V_1}^{V_2}dV=p(V_2-V_1)$
The problem gives T rather than p and V, so use the ideal gas law to rewrite the expression for W.
We know that pV = nRT, so $pV_1=nRT_1$ and $pV_2=nRT_2$
subtracting the two equations gives
$p(V_2-V_1)=nR(T_2-T_1)$
W = $nR(T_2-T_1)$
W = (2.00 mol)(8.3145 J/mol.K)(107 $^0$C - 27$^0$C)
W = +1330 J
The gas expands when heated and does positive work.
Q#19.2
Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. The initial temperature of the gas is 27.0$^0C$ and the pressure is constant. As part of a machine design project, calculate the final temperature of the gas after it has done 2.40 $\times 10^3$ J of work.
Answer:
At constant pressure, W = pΔV = nRΔT. Since the gas is doing work, it must be expanding, so ΔV is positive, which means that ΔT must also be positive.
R = 8.3145 J/mol.K, ΔT has the same numerical value in kelvins and in C°.
$\Delta T = \frac{W}{nR}=\frac{2.40 \times 10^3 \ J}{(6 \ mol)(8.3145 \ J/mol.K)}$
$\Delta T = 48.1 \ K$
$\Delta T_K = \Delta T_C$ and
$T_2=27.0^0C + 48.1^0C= 75.1^0C$
When W > 0 the gas expands. When p is constant and V increases, T increases.
Q#19.3
Two moles of an ideal gas are compressed in a cylinder at a constant temperature 65 $^0$C of until the original pressure has tripled. (a) Sketch a pV-diagram for this process. (b) Calculate the amount of work done.
Answer:
For an isothermal process, we use
W = nRT ln $\left(\frac{p_1}{p_2}\right)$
pV = nRT says V decreases when p increases and T is constant.
T = 65.0 + 273.15 = 338.15 K and $p_2=3p_1$
(a) The pV-diagram is sketched in Figure 19.3.
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| Fig.19.3 |
W = -6180 J
Since V decreases, W is negative.
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