a constant mass of gas maintained at constant pressure has a volume of 2.5×10⁴ at 0°c and 3.15×10⁴ at 100°c and 6.52×10⁴ at normal boiling point of sulphur the temperature corresponding to the boiling point of sulphur is

Question

a constant mass of gas maintained at constant pressure has a volume of 2.5×10⁴ at 0°c and 3.15×10⁴ at 100°c and 6.52×10⁴ at normal boiling point of sulphur the temperature corresponding to the boiling point of sulphur is

Asked on March 29, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-03-29T00:42:09.280Z

Step 1: Understand the relationship between volume and temperature. For a constant mass of gas maintained at constant pressure, the volume ($V$) is linearly related to the Celsius temperature ($T_C$). This relationship can be expressed as $V = aT_C + b$, where $a$ and $b$ are constants. Step 2: Use the given data points to find the constants $a$ and $b$. We have two known points: 1. At $T_C = 0^\circ C$, $V = 2.5 \times 10^4$. 2. At $T_C = 100^\circ C$, $V = 3.15 \times 10^4$. Substitute the first point into the equation: $$ 2.5 \times 10^4 = a(0) + b $$ $$ b = 2.5 \times 10^4 $$ Substitute the second point and the value of $b$ into the equation: $$ 3.15 \times 10^4 = a(100) + 2.5 \times 10^4 $$ $$ 100a = 3.15 \times 10^4 - 2.5 \times 10^4 $$ $$ 100a = (3.15 - 2.5) \times 10^4 $$ $$ 100a = 0.65 \times 10^4 $$ $$ a = \frac{0.65 \times 10^4}{100} $$ $$ a = 0.65 \times 10^2 $$ $$ a = 65 $$ So, the relationship between volume and temperature is: $$ V = 65T_C + 2.5 \times 10^4 $$ Step 3: Calculate the temperature corresponding to the boiling point of sulfur. We are given that the volume at the normal boiling point of sulfur is $V = 6.52 \times 10^4$. Let this temperature be $T_S$. Substitute this volume into the derived equation: $$ 6.52 \times 10^4 = 65T_S + 2.5 \times 10^4 $$ $$ 65T_S = 6.52 \times 10^4 - 2.5 \times 10^4 $$ $$ 65T_S = (6.52 - 2.5) \times 10^4 $$ $$ 65T_S = 4.02 \times 10^4 $$ $$ T_S = \frac{4.02 \times 10^4}{65} $$ $$ T_S = \frac{40200}{65} $$ $$ T_S \approx 618.4615 $$ Rounding to three significant figures, consistent with the input data: $$ T_S \approx 618^\circ C $$ The temperature corresponding to the boiling point of sulfur is $\boxed{618^\circ C}$.

Accepted Answer · 2026-03-29T00:42:09.280Z

Step 1: Understand the relationship between volume and temperature. For a constant mass of gas maintained at constant pressure, the volume ($V$) is linearly related to the Celsius temperature ($T_C$). This relationship can be expressed as $V = aT_C + b$, where $a$ and $b$ are constants. Step 2: Use the given data points to find the constants $a$ and $b$. We have two known points: 1. At $T_C = 0^\circ C$, $V = 2.5 \times 10^4$. 2. At $T_C = 100^\circ C$, $V = 3.15 \times 10^4$. Substitute the first point into the equation: $$ 2.5 \times 10^4 = a(0) + b $$ $$ b = 2.5 \times 10^4 $$ Substitute the second point and the value of $b$ into the equation: $$ 3.15 \times 10^4 = a(100) + 2.5 \times 10^4 $$ $$ 100a = 3.15 \times 10^4 - 2.5 \times 10^4 $$ $$ 100a = (3.15 - 2.5) \times 10^4 $$ $$ 100a = 0.65 \times 10^4 $$ $$ a = \frac{0.65 \times 10^4}{100} $$ $$ a = 0.65 \times 10^2 $$ $$ a = 65 $$ So, the relationship between volume and temperature is: $$ V = 65T_C + 2.5 \times 10^4 $$ Step 3: Calculate the temperature corresponding to the boiling point of sulfur. We are given that the volume at the normal boiling point of sulfur is $V = 6.52 \times 10^4$. Let this temperature be $T_S$. Substitute this volume into the derived equation: $$ 6.52 \times 10^4 = 65T_S + 2.5 \times 10^4 $$ $$ 65T_S = 6.52 \times 10^4 - 2.5 \times 10^4 $$ $$ 65T_S = (6.52 - 2.5) \times 10^4 $$ $$ 65T_S = 4.02 \times 10^4 $$ $$ T_S = \frac{4.02 \times 10^4}{65} $$ $$ T_S = \frac{40200}{65} $$ $$ T_S \approx 618.4615 $$ Rounding to three significant figures, consistent with the input data: $$ T_S \approx 618^\circ C $$ The temperature corresponding to the boiling point of sulfur is $\boxed{618^\circ C}$.