Q#19.24
Three moles of an ideal monatomic gas expands at a constant pressure of 2.50 atm; the volume of the gas changes from 3.2 $\times 10^{-2} \ m^3$ to 4.5 $\times 10^{-2} \ m^3$.
(a) Calculate the initial and final temperatures of the gas.
(b) Calculate the amount of work the gas does in expanding.
(c) Calculate the amount of heat added to the gas. (d) Calculate the change in internal energy of the gas.
Answer:
Apply pV = nRT to calculate T. For this constant pressure process, W = p$\Delta V$, Q = n$C_p\Delta T$
Use $\Delta U = Q - W$ to relate Q, W and ΔU.
Given: p = 2.50 atm = 2.53 $\times 10^5$ Pa.
For a monatomic ideal gas, $C_v$ = 12.47 J/mol.K and $C_p$ = 20.78 J/mol.K
(a) $T_1=\frac{pV_1}{nR}$
$T_1= \frac{(2.53 \times 10^5 \ Pa)(3.20 \times 10^{-2} \ m^3)}{(3.00 \ mol)(8.3145 \ J/mol.K)}$ = 325 K
(b) W = p$\Delta V$ = (2.53 $\times 10^5$ Pa)(4.5 $\times 10^{-2} \ m^3-$ 3.2 $\times 10^{-2} \ m^3$)
W = 3.29 $\times 10^3$ J
(c) Q = n$C_p\Delta T$ = (3.00 mol)(20.78 J/mol.K)(456 K - 325 K) = 8.17 $\times 10^3$ J
(d) $\Delta U = Q - W$ = 8.17 $\times 10^3$ J - 3.29 $\times 10^3$ J
$\Delta U = 4.88 \times 10^3$ J
Q#19.25
A cylinder with a movable piston contains 3.00 mol of $N_2$ gas (assumed to behave like an ideal gas).
(a) The $N_2$ is heated at constant volume until 1557 J of heat have been added. Calculate the change in temperature.
(b) Suppose the same amount of heat is added to the $N_2$ but this time the gas is allowed to expand while remaining at constant pressure. Calculate the temperature change.
(c) In which case, (a) or (b), is the final internal energy of the $N_2$ higher? How do you know? What accounts for the difference between the two cases?
Answer:
For constant volume Q = n$C_v\Delta T$
For constant pressure Q = n$C_p\Delta T$
For any process of an ideal gas $\Delta U = nC_v \Delta T$
Given: For $N_2$, $C_v$ = 20.78 J/mol.K and $C_p$ = 29.07 J/mol.K
Heat is added, so Q is positive and Q = +1557 K.
(a) $\Delta T = \frac{Q}{nC_v} = \frac{1557 \ J}{(3.00 \ mol)(20.76 \ J/mol.K)} = +25.0 \ K$
(b) $\Delta T = \frac{Q}{nC_v} = \frac{1557 \ J}{(3.00 \ mol)(29.07 \ J/mol.K)} = +17.9 \ K$
(c) $\Delta U = nC_v \Delta T$ for either process, so ΔU is larger when ΔT is larger. The final internal energy is larger for the constant volume process in (a).
For constant volume W = 0 and all the energy added as heat stays in the gas as internal energy. For the constant pressure process the gas expands and W > 0. Part of the energy added as heat leaves the gas as expansion work done by the gas.
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