Новая Школа - онлайн-школа подготовки к ЕГЭ
Задача · 8 класс Задача №3706081
А
Аноним
02 октября 2026 02:47 · публичный вопрос
А
Аноним

Use the graph to answer the question

Ответ нейросети

Brief answers

Using the approximate graph values \(P_{\min}\approx99\), \(P_{\max}\approx115\), \(T_{\min}=25^\circ\text{C}\), \(T_{\max}=95^\circ\text{C}\):

  1. As temperature increases, pressure increases.
  2. Gay-Lussac’s Law (also called Amontons’s Law).
  3. \[ \frac{P_{\max}}{P_{\min}}\approx\frac{115}{99}\approx1.16 \] \[ \frac{T_{\max}}{T_{\min}}=\frac{95}{25}=3.80 \]
  4. According to the data, pressure would reach zero at approximately \(-400^\circ\text{C}\), about \(-408^\circ\text{C}\) using the endpoint values.
  5. Pressure of zero would mean that gas particles make no collisions with the container walls. In the ideal-gas model, their motion would have essentially stopped.
  6. \[ 95^\circ\text{C}=95+273=\boxed{368\text{ K}} \] \[ 25^\circ\text{C}=25+273=\boxed{298\text{ K}} \]
  7. \[ \frac{368}{298}\approx\boxed{1.24} \] This is reasonably close to the pressure ratio \(1.16\). The comparison works with Kelvin, not Celsius.
  8. Paste the corrected graph: temperature on the \(x\)-axis, pressure on the \(y\)-axis.

Key reasoning

For Gay-Lussac’s Law, volume is constant and

\[ P\propto T_{\text{K}}. \]

Using the approximate endpoint data:

\[ m=\frac{115-99}{95-25}=\frac{16}{70}\approx0.229. \]

The line is approximately

\[ P=99+0.229(T-25). \]

For zero pressure:

\[ 0=99+0.229(T-25), \]

so

\[ T\approx-408^\circ\text{C}. \]

Important: in the photograph, pressure appears to be on the horizontal axis and temperature on the vertical axis. For question 4, swap the axes so that \(T(^{\circ}\text{C})\) is on \(x\) and \(P\) is on \(y\).

Can you send the original data table for an exact intercept?

Обсуждение

2 сообщения
А
Аноним

Can you make it like an 9th grader wrote it and like it’s wrote on a computer

Нейросеть

Answers:

  1. When the temperature increases, the pressure also increases.

  2. This graph shows Gay-Lussac’s Law. It explains that pressure and temperature increase together when the volume stays the same.

  3. Pressure ratio:

115 ÷ 99 ≈ 1.16

Temperature ratio:

95 ÷ 25 = 3.8

  1. If the graph continued, the pressure would reach zero at about -400°C, or around -408°C.

  2. A pressure of zero would mean that the gas particles are not hitting the walls of the container. In the ideal gas model, the particles would have almost stopped moving.

  3. Converting the temperatures to Kelvin:

95°C + 273 = 368 K

25°C + 273 = 298 K

  1. Temperature ratio:

368 ÷ 298 ≈ 1.24

This is fairly close to the pressure ratio of 1.16. The temperatures need to be measured in Kelvin instead of Celsius.

  1. The corrected graph should have temperature in °C on the x-axis and pressure on the y-axis.