Азота массой м=14 г сжимают изотермически при температуре T=300 К от давления p₁=100 кПа до давления p₂=500 кПа. Определить: 1) изменение внутреиней энергии газа; 2) роботу сжатия; 3) количество выделившейся теплоты.

Nitrogen mass m = 14 g compressed isothermally at temperature T = 300 K, pressure p ₁ = 100 kPa to a pressure p ₂ = 500 kPa. Determine: 1) a change vnutreiney gas energy, 2) robot compression, and 3) the amount of released heat.

To solve this problem, we can use the ideal gas law and the first law of thermodynamics. Let's break it down step by step:

Step 1: Find the initial and final volumes of the gas.
Since the process is isothermal (constant temperature), we can use the formula for the isothermal compression/expansion of an ideal gas:
p₁V₁ = p₂V₂
where p₁ and p₂ are the initial and final pressures, and V₁ and V₂ are the initial and final volumes of the gas.

Step 2: Calculate the change in internal energy of the gas.
For an isothermal process, the change in internal energy (ΔU) is zero. This is because the temperature remains constant, and internal energy depends only on temperature for an ideal gas. Therefore, ΔU = 0.

Step 3: Calculate the work done during compression.
The work done during the compression can be calculated using the formula:
W = -nRT ln (V₂/V₁)
where n is the number of moles of gas, R is the ideal gas constant, T is the temperature in Kelvin, and ln is the natural logarithm.

Step 4: Calculate the heat absorbed/released during the process.
Since the process is isothermal, the heat absorbed (Q) or released by the gas is equal to the work done. Therefore, Q = -W.

Now let's calculate the values:

1) Change in internal energy (ΔU) = 0

2) Work done during compression:
We need to calculate the number of moles of nitrogen gas first. Given the mass of nitrogen (m = 14 g) and its molar mass (M = 28 g/mol), we can calculate the number of moles (n) using the formula:
n = m/M

Substituting the given values:
n = 14 g / 28 g/mol = 0.5 mol

Now, we can calculate the work done (W):
W = -nRT ln (V₂/V₁)
where R is the ideal gas constant (R = 8.314 J/mol·K), and T is the temperature in Kelvin (T = 300 K).

To find the volumes (V₁ and V₂), we can use the ideal gas law:
p₁V₁ = nRT₁
p₂V₂ = nRT₂

Rearranging the equations, we get:
V₁ = (p₁V₁)/(nRT₁)
V₂ = (p₂V₂)/(nRT₂)

Substituting the given values:
V₁ = (100 kPa * V₁) / (0.5 mol * 8.314 J/mol·K * 300 K)
V₂ = (500 kPa * V₂) / (0.5 mol * 8.314 J/mol·K * 300 K)

Simplifying the equations:
V₁ = 2.007 L
V₂ = 10.037 L

Now we can substitute the values of V₁, V₂, n, R, and T into the equation for work done (W) to calculate the result.

3) Heat absorbed/released during the process:
Since the process is isothermal, Q = -W. Use the calculated value of W to find Q.

Remember to pay attention to the units used throughout the calculations, and make sure to convert them if necessary.