CLEANED: A 12 M Long Bar When Heated From 20°C To 70°C ✓ Solved

CLEANED A 12 M Long Bar When Heated From 20 C To 70 C Becomes 12008 M

CLEANED: A 12 M Long Bar When Heated From 20 C To 70 C Becomes 12008 M

Calculate the linear expansion coefficient of the bar given that a 12-meter long bar, when heated from 20°C to 70°C, becomes 12008 meters long. Additionally, determine the volume change of a steel cube with side length 4 meters heated by 120 K, given its linear expansion coefficient of 11×10-6 °K-1. Using the ideal gas law PV = nRT with R = 8.3 J/(°K·mol), compute the volume of one mol of air at 20°C and 105 Pa. Finally, convert temperatures T₁ = 40°C and T₂ = 80°C into Kelvin, then calculate the differences: T₁ - T₂, the ratio T₁ / T₂, and the natural logarithm ln(T₁ / T₂).

Sample Paper For Above instruction

Introduction

Thermal expansion and ideal gas laws are fundamental concepts in physics that describe the behavior of materials and gases under temperature and pressure changes. These principles have widespread applications, from engineering to environmental science. In this paper, we analyze the linear expansion coefficient of a metallic bar, compute the volumetric change in a steel cube upon heating, determine the volume of a mole of gas under specific conditions, and explore temperature conversions and their mathematical relationships. By understanding these concepts, we can better predict the physical behavior of materials and gases in real-world scenarios.

Calculation of the Linear Expansion Coefficient

The problem states that a 12-meter-long bar, when heated from 20°C to 70°C, expands to 12008 meters. This appears to be a typographical or data entry error, as expanding from 12 meters to 12008 meters would be physically unreasonable. Assuming the intended value was 12.008 meters—the usual form of such problems—the change in length ΔL is 0.008 meters. The initial length L₀ is 12 meters, the temperature change ΔT is 50°C (from 20°C to 70°C). The linear expansion coefficient α is given by:

α = ΔL / (L₀ × ΔT) = 0.008 m / (12 m × 50°C) = 0.008 / 600 = 1.333×10-5 °C-1.

This coefficient indicates how much a material expands per degree increase in temperature. The value aligns with typical metallic expansion coefficients, suggesting consistent material behavior.

Volume Change in Steel Cube

Given the linear expansion coefficient of steel as 11×10-6 °K-1, the volumetric expansion coefficient β is approximately three times α:

β ≈ 3α = 3 × 11×10-6 = 33×10-6 or 3.3×10-5 °K-1.

For a cube with side length L = 4 meters, the change in temperature ΔT = 120 K. The volume expansion ΔV is given by:

ΔV = β × V₀ × ΔT, where V₀ = L³ = 4³ = 64 m³.

Calculating ΔV:

ΔV = 3.3×10-5 × 64 m³ × 120 K ≈ 3.3×10-5 × 7680 ≈ 0.25344 m³.

Thus, the new volume Vnew ≈ V₀ + ΔV ≈ 64 + 0.25344 ≈ 64.25344 m³. The steel cube will expand by approximately 0.25344 cubic meters when heated by 120 K.

Volume of One Mole of Gas

Applying the ideal gas law PV = nRT, with n = 1 mol, P = 105 Pa, T = 20°C = 293.15 K, and R = 8.3 J/(°K·mol):

V = nRT / P = (1 mol) × 8.3 J/(°K·mol) × 293.15 K / 105 Pa.

Calculating numerator:

8.3 × 293.15 ≈ 2433.65 J

Therefore:

V ≈ 2433.65 J / 105 Pa = 231.75 m3.

This large volume reflects the low pressure and typical room temperature conditions for a mol of gas.

Temperature Conversion and Mathematical Calculations

Converting T₁ = 40°C and T₂ = 80°C into Kelvin:

T₁ = 40 + 273.15 = 313.15 K.

T₂ = 80 + 273.15 = 353.15 K.

Calculating T₁ - T₂:

313.15 - 353.15 = -40 K.

Calculating T₁ / T₂:

313.15 / 353.15 ≈ 0.887.

Calculating the natural logarithm ln(T₁ / T₂):

ln(0.887) ≈ -0.120.

These calculations demonstrate the relationships between temperatures in Kelvin and their ratios and logarithms, which are essential in thermodynamics and statistical mechanics.

Conclusion

The analysis of thermal expansion coefficients reveals the thermal behavior of metals and gases. Accurate calculations of volume expansion and gas volume under specific conditions help in designing safe and efficient engineering systems. Temperature conversions between Celsius and Kelvin and related mathematical functions are vital in scientific calculations. Together, these concepts underscore the importance of fundamental physics principles in real-world applications and scientific research.

References

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