Question

При Т=298 К 1×10-1 кг N2 адиабатически расширяют от 14×10-4 м3 до 14×10-2 м3. Определите конечную температуру Т, работу расширения, DU и DН.

Answer 1

Ответ:

47 К,

5215 Дж,

-5215 Дж,

-52208 Дж.

Объяснение:

Для определения конечной температуры воспользуемся уравнением адиабаты:

$TV^\gamma ^-^1=const$ , где

γ=1,4 - коэффициент Пуассона для двухатомного газа,

Для двух последовательных состояний запишем:

$T_1V_1^\gamma ^-^1=T_2V_2^\gamma ^-^1$

откуда:

$T_2=T_1\frac{V_1^\gamma ^-^1}{V_2^\gamma ^-^1} =298\frac{(14*10^-^4)^0^.^4}{(14*10^-^2)^0^.^4} =47$ К

Работу расширения определим по формуле:

$A=\frac{R*(T_1-T_2)}{\gamma -1} =\frac{8.31*(298-47)}{1.4-1}=5215$ Дж

Изменение внутренней энергии равно работе со знаком минус по первому началу термодинамики:

$\Delta U=-A=-5215$ Дж

А изменение энтальпии рассчитаем по формуле:

$\Delta H=c_p*(T_2-T_1)$, где

Cp - изобарная теплоемкость, равная

$c_p=\frac{i+2}{2} \frac{m}{M}R=\frac{5+2}{2} \frac{0.1}{0.014}*8.31=208$ Дж/К

$\Delta H=208(47-298)=-52208$ Дж.

Answer 2

#### Problem Statement We are given the following information: - Initial temperature (T): 298 K - Initial volume (V1): 14×10^(-4) m^3 - Final volume (V2): 14×10^(-2) m^3 We need to determine the final temperature (T), the work done during expansion (W), the change in internal energy (ΔU), and the change in enthalpy (ΔH). #### Solution To solve this problem, we can use the ideal gas law and the first law of thermodynamics. #### Calculating the Final Temperature (T) The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Since the process is adiabatic (no heat exchange), we can assume that the number of moles (n) and the gas constant (R) remain constant. Therefore, we can write: P1V1^γ = P2V2^γ Where P1 and P2 are the initial and final pressures, and γ is the heat capacity ratio (Cp/Cv) for the gas. Since the process is adiabatic, we can use the adiabatic index for nitrogen gas (N2), which is approximately 1.4. We can rearrange the equation to solve for the final temperature (T): T = (P2V2^γ) / (P1V1^γ) Let's calculate the final temperature using the given values. **Final Temperature (T):** T = (P2V2^γ) / (P1V1^γ) T = (P2 / P1) * (V2 / V1)^γ T = (P2 / P1) * (14×10^(-2) / 14×10^(-4))^1.4 T = (P2 / P1) * (10^2)^1.4 T = (P2 / P1) * 100^1.4 Now, we need to find the values of P1 and P2. Unfortunately, we don't have this information available in the given question. Therefore, we cannot calculate the final temperature (T) without the pressure values. #### Calculating the Work Done during Expansion (W) The work done during expansion can be calculated using the formula: W = -PΔV Where P is the pressure and ΔV is the change in volume. Since the process is adiabatic, we can assume that the pressure (P) remains constant. Therefore, we can write: W = -P(V2 - V1) Let's calculate the work done during expansion using the given values. **Work Done during Expansion (W):** W = -P(V2 - V1) W = -P(14×10^(-2) - 14×10^(-4)) W = -P(14×10^(-2) - 14×10^(-4)) Now, we need to find the value of P. Unfortunately, we don't have this information available in the given question. Therefore, we cannot calculate the work done during expansion (W) without the pressure value. #### Calculating ΔU and ΔH Since we don't have the values of pressure (P1 and P2), we cannot calculate the change in internal energy (ΔU) and the change in enthalpy (ΔH) without this information. #### Conclusion In conclusion, we cannot calculate the final temperature (T), the work done during expansion (W), the change in internal energy (ΔU), and the change in enthalpy (ΔH) without the values of pressure (P1 and P2).