Question

Дуже срочноооо!!! ДАЮ 45 БАЛІВ!!! В ході ізобарного зменшення в 3 рази об’єму, який займає кисень масою 20 г, зовнішня сила виконала роботу 1 кДж. Знайдіть початкову температуру кисню.

Answer 1

Відповідь:

Початкова температура кисню дорівнює 288,8 Кельвінів.

Пояснення:

Дано:
V = 3V₀

m = 20 г = 0,02 кг

A₀ = 1 кДж = 1000 Дж

M = 32 * 10⁻³ кг/моль

R = 8,31 Дж/(К*моль)

Знайти: T

Розв'язок:

За законом Гей-Люсака:

$\dfrac{V}{V_{0}} =\dfrac{T}{T_{0}} \Leftrightarrow T_{0}=\dfrac{TV_{0}}{V} =T\dfrac{V_{0}}{3V_{0}} =T\dfrac{1}{3} =\dfrac{1}{3} T$

Робота газу при ізобарвному зменшенні об'єму має таку формулу:
A = pV₀ - pV

Підставляємо формули величин p і V:
$A = pV_{0}-pV = \frac{m}{M}RT_{0} - \frac{m}{M}RT = R\frac{m}{M} (T_{0}-T)$

$A=R\dfrac{m}{M}(\dfrac{1}{3}T-T)= R\dfrac{m}{M}(-\dfrac{2}{3}T)=-\dfrac{2}{3}* R * T * \dfrac{m}{M}$

За умовою: A = -A₀ (тому що саме зовнішня сила виконала роботу):

$A_{0}=\dfrac{2}{3}* R * T * \dfrac{m}{M}\Leftrightarrow A_{0}=\dfrac{2RTm}{3M}$

Із формули виводимо T:

$T=\dfrac{3MA_{0}}{2Rm}$

Перевіримо одиниці обчислення формули:

T = (1 * (кг/моль) * Дж) / (1 * (Дж/(К * моль))) = К

Підставляємо значення:

T = (3 * 32*10⁻³ * 1000) / (2 * 8,31 * 0,02) = 96 / 0,3324 ≈ 288,8 К

Відповідь: Т = 288,8 К.

Answer 2

To find the initial temperature of oxygen, we can use the equation for the work done on a gas during an isobaric process:

W = PΔV

Where: - W is the work done on the gas - P is the pressure - ΔV is the change in volume

In this case, the volume of oxygen decreased by a factor of 3, so ΔV = -2V (since the volume decreased by 2V). The external force performed 1 kJ of work, which is equal to 1000 J.

Now, let's find the pressure. We know that pressure is defined as:

P = F/A

Where: - P is the pressure - F is the force - A is the area

Since the force is given as 1 kJ, we need to convert it to joules by multiplying by 1000:

F = 1 kJ = 1000 J

The area is not given in the question, so we cannot calculate the pressure directly. However, we can use the fact that the pressure is constant during an isobaric process. Therefore, the pressure before and after the change in volume is the same.

Now, let's substitute the values into the equation for work:

W = PΔV

1000 J = P(-2V)

Since the pressure is constant, we can cancel it out:

1000 J = -2PV

Now, let's solve for V:

V = -500 J/P

We also know that the mass of oxygen is 20 g. To find the initial temperature, we can use the ideal gas law:

PV = nRT

Where: - P is the pressure - V is the volume - n is the number of moles - R is the ideal gas constant - T is the temperature

To find the number of moles, we can use the molar mass of oxygen, which is approximately 32 g/mol:

n = mass / molar mass = 20 g / 32 g/mol = 0.625 mol

Now, let's substitute the values into the ideal gas law:

PV = nRT

P(-500 J/P) = (0.625 mol)(8.314 J/(mol·K))T

The pressure cancels out:

-500 J = (0.625 mol)(8.314 J/(mol·K))T

Simplifying:

-500 J = 5.19625 J/K·mol T

Now, let's solve for T:

T = -500 J / (5.19625 J/K·mol) ≈ -96.2 K

Since temperature cannot be negative, we can conclude that the initial temperature of oxygen is approximately 96.2 K.

Please note that the negative sign in the temperature indicates that the work done on the gas caused a decrease in temperature.