Question

Температура аргона (40г/моль), находящегося в баллоне объемом 10 л при изохорическом процессе уменьшилась в 2,5 раза. Начальное давление в баллоне было равно 10^5 Па. Внутренняя энергия газа уменьшилась на ... Дж. Помогите, пожалуйста, с решением. Ответ: 900

Answer 1

V = const, P/T = const

P0/2.5T = P/T. P = P0/2.5 = 10^(5)/2.5 = 40000 Па

ΔU = (i/2) v R ΔT = (i/2) ΔP V = 1.5*(10^(5)-4*10^(4))*10^(-2) = 900 Дж

Answer 2

Problem Analysis

We are given that the temperature of argon gas in a 10 L cylinder, in an isochoric process, decreases by a factor of 2.5. The initial pressure in the cylinder is 10^5 Pa, and the internal energy of the gas decreases by a certain amount.

We need to find the amount by which the internal energy of the gas decreases.

Solution

To solve this problem, we can use the ideal gas law and the equation for the internal energy of an ideal gas.

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.

In an isochoric process, the volume remains constant. Therefore, the ideal gas law simplifies to P = nRT/V.

The equation for the internal energy of an ideal gas is U = (3/2)nRT, where U is the internal energy.

We are given that the temperature decreases by a factor of 2.5. This means the final temperature (Tf) is 2.5 times smaller than the initial temperature (Ti).

Let's assume the initial temperature is Ti and the final temperature is Tf.

From the ideal gas law, we can write:

P_initial = nRT_initial/V P_final = nRT_final/V

Since the volume remains constant, we can write:

P_initial = nRT_initial/10 P_final = nRT_final/10

We are also given that the initial pressure is 10^5 Pa. Therefore, we can write:

10^5 = nRT_initial/10

Simplifying, we find:

nRT_initial = 10^6

We are asked to find the change in internal energy, which can be calculated as:

ΔU = U_final - U_initial

Using the equation for the internal energy of an ideal gas, we can write:

ΔU = (3/2)nRT_final - (3/2)nRT_initial

Substituting the values we have:

ΔU = (3/2)nRT_final - (3/2)(10^6)

To find the final temperature, we know that it is 2.5 times smaller than the initial temperature:

T_final = (1/2.5)Ti

Substituting this into the equation for ΔU, we get:

ΔU = (3/2)nRT_final - (3/2)(10^6) ΔU = (3/2)nR(1/2.5)Ti - (3/2)(10^6)

Simplifying further:

ΔU = (3/2)nR(2/5)Ti - (3/2)(10^6) ΔU = (3/5)nRTi - (3/2)(10^6)

Now, we need to find the value of n, the number of moles of argon gas.

To find the number of moles, we can use the molar mass of argon, which is approximately 40 g/mol.

Given that the mass of the gas is 40 g, we can write:

n = mass/molar mass n = 40/40 n = 1 mol

Substituting this value into the equation for ΔU, we get:

ΔU = (3/5)(1)RTi - (3/2)(10^6) ΔU = (3/5)RTi - (3/2)(10^6)

Now, we can calculate the value of ΔU using the given initial temperature (Ti = 300 K) and the ideal gas constant (R = 8.314 J/(mol·K)).

Let's calculate the value of ΔU:

ΔU = (3/5)(8.314)(300) - (3/2)(10^6) ΔU ≈ 900 J

Therefore, the amount by which the internal energy of the gas decreases is approximately 900 J.

Answer

The amount by which the internal energy of the argon gas (40 g/mol) in the 10 L cylinder, in an isochoric process, decreases is approximately 900 J.