Q#19.48
The graph in Fig. P19.48 shows a pV-diagram for 3.25 moles of ideal helium (He) gas. Part ca of this process is isothermal.
(a) Find the pressure of the He at point a.
(b) Find the temperature of the He at points a, b, and c.
(c) How much heat entered or left the He during segments ab, bc, and ca? In each segment, did the heat enter or leave?
(d) By how much did the internal energy of the He change from a to b, from b to c, and from c to a? Indicate whether this energy increased or decreased.
Answer:
Segment ab is isobaric, bc is isochoric, and ca is isothermal.
He is a monatomic gas so $C_v=\frac{3}{2}R$ and $C_p=\frac{5}{2}R$. For any process of an ideal gas, $\Delta U = nC_v\Delta T$.
For an isothermal process of an ideal gas, $\Delta U = 0$ so $Q = W = nRT \ ln(V_2/V_1)$
(a) Apply pV = nRT to states a and c. $T_a=T_c$ so nRT is constant and $p_aV_a=p_cV_c$
$p_a=p_c \times \frac{V_c}{V_a}$
$p_a=(2.0 \times 10^5 \ Pa) \times \frac{0.040 \ m^3}{0.010 \ m^3}=8.0 \times 10^5 \ Pa$
(b) $T_a=\frac{p_aV_a}{nR}$
$T_a=\frac{(8.0 \times 10^5 \ Pa)(0.010 \ m^3)}{(3.25 \ mol)(8.3145 \ J/mol.K)}=296 \ K$
$T_b=\frac{p_bV_b}{nR}$
$T_b=\frac{(8.0 \times 10^5 \ Pa)(0.040 \ m^3)}{(3.25 \ mol)(8.3145 \ J/mol.K)}=1184 \ K$
$T_c=\frac{p_cV_c}{nR}$
$T_a=\frac{(2.0 \times 10^5 \ Pa)(0.040 \ m^3)}{(3.25 \ mol)(8.3145 \ J/mol.K)}=296 \ K=T_a$
(c) For ab:
Q = $nC_p\Delta T$
Q = $(3.25 \ mol)(\frac{5}{2})(8.3145 \ J/mol.K)(1184 \ K-296 \ K)=6.00 \times 10^4 \ J$, heat enters the gas.
For bc:
Q = $nC_v\Delta T$
Q = $(3.25 \ mol)(\frac{3}{2})(8.3145 \ J/mol.K)(296 \ K-1184 \ K)=-3.60 \times 10^4 \ J$, heat leaves the gas.
For ca:
Q = $nRT \ ln(V_a/V_c)$
Q = $(3.25 \ mol)(8.3145 \ J/mol.K)(296 \ K) \ ln \left(\frac{0.010 \ m^3}{0.040 \ m^3}\right)=-1.10 \times 10^4 \ J$, heat leaves the gas.
(d) For ab:
$\Delta U =nC_v\Delta T$
$\Delta U = (3.25 \ mol)(\frac{3}{2})(8.3145 \ J/mol.K)(1184 \ K-296 \ K)=3.60 \times 10^4 \ J$, the internal energy increased.
For bc:
$\Delta U =nC_v\Delta T$
$\Delta U = (3.25 \ mol)(\frac{3}{2})(8.3145 \ J/mol.K)(296 \ K -1184 \ K)=-3.60 \times 10^4 \ J$, the internal energy decreased.
For ca:
$\Delta T = 0$ so $\Delta U = 0$
As we saw in (d), for any closed path on a pV diagram, ΔU = 0 because we are back at the same values of P, V, and T.
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