Thermal Physics Alıştırma Soruları — JAMB UTME — Sayfa 9

Soru 161Soru →

A container made of a metal with a linear expansivity of $2.0 \times 10^{-5}\text{ K}^{-1}$ has a volume of $1000\text{ cm}^3$ at $20^\circ\text{C}$ . It is completely filled with a liquid at this temperature. When the system is heated to $120^\circ\text{C}$ , $40\text{ cm}^3$ of the liquid overflows. If the real cubic expansivity of the liquid is expressed as $X \times 10^{-4}\text{ K}^{-1}$ , what is the value of $X$ ?

Cevabı ve açıklamayı göster

Cevap: 4.6

Cevap

The value of

X

is 4.6.

The real cubic expansivity of a liquid is given by

\gamma_r = \gamma_a + \gamma_v

. Calculating apparent expansivity gives

\gamma_a = \frac{40}{1000 \times 100} = 4.0 \times 10^{-4}\text{ K}^{-1}

. The vessel's volume expansivity is

\gamma_v = 3 \times 2.0 \times 10^{-5} = 0.6 \times 10^{-4}\text{ K}^{-1}

. Adding them together yields

\gamma_r = 4.6 \times 10^{-4}\text{ K}^{-1}

, so

X = 4.6

.

Adım Adım Çözüm

1

Calculate the temperature change

\Delta T

\Delta T = 120^\circ\text{C} - 20^\circ\text{C} = 100\text{ K}

The thermal expansion is driven by the change in temperature.

2

Calculate the apparent cubic expansivity

\gamma_a

\gamma_a = \frac{40\text{ cm}^3}{1000\text{ cm}^3 \times 100\text{ K}} = 4.0 \times 10^{-4}\text{ K}^{-1}

Apparent expansivity is defined as the fraction of initial volume overflowed per degree rise in temperature.

3

Calculate the volume expansivity of the container

\gamma_v

\gamma_v = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1} = 0.6 \times 10^{-4}\text{ K}^{-1}

Cubic expansivity of an isotropic solid container is three times its linear expansivity.

4

Sum apparent expansivity and vessel cubic expansivity to find real cubic expansivity

\gamma_r

\gamma_r = \gamma_a + \gamma_v = 4.0 \times 10^{-4} + 0.6 \times 10^{-4} = 4.6 \times 10^{-4}\text{ K}^{-1}

Real expansivity accounts for both the apparent expansion of the liquid and the expansion of the container.

Anahtar Kavram

Relationship between real cubic expansivity, apparent cubic expansivity, and vessel expansivity

Soru 162Soru →

A rectangular metallic sheet with a linear expansivity of $1.8 \times 10^{-5}\text{ K}^{-1}$ experiences a temperature rise of $50\text{ K}$ . If the increase in its surface area is $0.90\text{ cm}^2$ , what was the initial surface area of the sheet in $\text{cm}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 500

Cevap

The initial surface area of the metallic sheet is

500\text{ cm}^2

.

The initial area is found by converting linear expansivity to area expansivity (\beta = 2\alpha = 3.6 \times 10^{-5}\text{ K}^{-1}) and substituting into the area expansion relation \Delta A = A_0 \beta \Delta T, giving A_0 = \frac{0.90}{3.6 \times 10^{-5} \times 50} = 500\text{ cm}^2$.

Adım Adım Çözüm

1

Calculate the area (superficial) expansivity (\beta)

\beta = 2\alpha = 2 \times 1.8 \times 10^{-5}\text{ K}^{-1} = 3.6 \times 10^{-5}\text{ K}^{-1}

Surface area expansion depends on area expansivity, which is twice the linear expansivity for an isotropic solid.

2

Formulate the thermal area expansion equation

\Delta A = A_0 \beta \Delta T

The fractional change in area is directly proportional to the area expansivity and the temperature change.

3

Rearrange the formula to solve for the initial surface area (A_0)

A_0 = \frac{\Delta A}{\beta \Delta T}

Isolating the required unknown quantity.

4

Substitute the known numerical values and compute

A_0 = \frac{0.90\text{ cm}^2}{(3.6 \times 10^{-5}\text{ K}^{-1})(50\text{ K})} = \frac{0.90}{1.8 \times 10^{-3}} = 500\text{ cm}^2

Evaluating the expression yields the exact initial surface area.

Anahtar Kavram

Relationship between Linear Expansivity and Area Expansivity

Alternatif Yöntem

Calculate fractional area expansion per kelvin: \beta = 2\alpha = 3.6 \times 10^{-5}\text{ K}^{-1}

. Total fractional expansion for

50\text{ K}

is

3.6 \times 10^{-5} \times 50 = 0.0018

. Then initial area

A_0 = \frac{0.90}{0.0018} = 500\text{ cm}^2$.

Tahmini Süre:1m 30s

Soru 163Soru →

A sealed cylinder in an industrial pneumatic lift contains $0.080\text{ m}^3$ of gas at an initial pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The piston compresses the gas to a final volume of $0.020\text{ m}^3$ , raising the pressure to $7.50 \times 10^5\text{ Pa}$ . What is the final temperature of the gas in degrees Celsius ( $^\circ\text{C}$ )?

Cevabı ve açıklamayı göster

Cevap: 102

Cevap

The final temperature of the gas is

102^\circ\text{C}

.

Using the combined gas law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

with absolute initial temperature

T_1 = 300\text{ K}

gives

T_2 = 375\text{ K}

. Subtracting

273

yields the final temperature of

102^\circ\text{C}

.

Adım Adım Çözüm

1

Convert initial temperature to absolute temperature (Kelvin)

T_1 = 27 + 273 = 300\text{ K}

Gas laws require thermodynamic temperature measured on the Kelvin scale.

2

Set up the Combined Gas Law equation

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Relates initial and final states when pressure, volume, and temperature all change for a fixed mass of gas.

3

Calculate the final absolute temperature

T_2

T_2 = 300 \times \frac{7.50 \times 10^5 \times 0.020}{1.50 \times 10^5 \times 0.080} = 375\text{ K}

Evaluating the ratio of

P_2 V_2

to

P_1 V_1

yields

1.25

.

4

Convert final temperature from Kelvin to Celsius

\theta_2 = 375 - 273 = 102^\circ\text{C}

The question explicitly requires the answer in degrees Celsius.

Anahtar Kavram

Combined Gas Law and thermodynamic temperature conversion

Soru 164Soru →

An aluminum electric cable suspended between two transmission poles has a length of $100\text{ m}$ at an initial morning temperature of $20^\circ\text{C}$ . By afternoon, the cable temperature increases to $50^\circ\text{C}$ . Given that the linear expansivity of aluminum is $2.3 \times 10^{-5}\text{ K}^{-1}$ , what is the increase in the length of the cable in centimeters?

Cevabı ve açıklamayı göster

Cevap: 6.9

Cevap

The increase in the length of the cable is

6.9\text{ cm}

.

Applying the formula for linear expansion

\Delta L = L_0 \alpha \Delta T

, where

L_0 = 100\text{ m}

,

\alpha = 2.3 \times 10^{-5}\text{ K}^{-1}

, and

\Delta T = 30\text{ K}

, gives

\Delta L = 0.069\text{ m}

. Converting this to centimeters yields

6.9\text{ cm}

.

Adım Adım Çözüm

1

Determine the temperature change

\Delta T = 30\text{ K}

Temperature change is the difference between final and initial temperatures:

50^\circ\text{C} - 20^\circ\text{C} = 30\text{ K}

.

2

Calculate expansion in meters using the linear expansivity formula

\Delta L = 0.069\text{ m}

\Delta L = L_0 \alpha \Delta T = 100 \times (2.3 \times 10^{-5}) \times 30 = 0.069\text{ m}.

3

Convert the change in length to centimeters

\Delta L = 6.9\text{ cm}

Since

1\text{ m} = 100\text{ cm}

, multiply

0.069\text{ m}

by

100

to obtain

6.9\text{ cm}

.

Anahtar Kavram

Linear expansivity of solid conductors

Tahmini Süre:1m 30s

Soru 165Soru →

A glass flask of volume $800\text{ cm}^3$ at $10^\circ\text{C}$ is completely filled with a liquid having a real volume expansivity of $5.0 \times 10^{-4}\text{ K}^{-1}$ . If the linear expansivity of the glass is $1.0 \times 10^{-5}\text{ K}^{-1}$ , what volume of the liquid (in $\text{cm}^3$ ) will overflow when the flask and its contents are heated to $60^\circ\text{C}$ ?

Cevabı ve açıklamayı göster

Cevap: 18.8

Cevap

18.8 cm^3

The volume of liquid that overflows is equal to the apparent increase in volume of the liquid. The apparent volume expansivity of the liquid $\gamma_a$ is given by $\gamma_a = \gamma_r - \gamma_v$, where $\gamma_r = 5.0 \times 10^{-4}\text{ K}^{-1}$ is the real expansivity of the liquid, and $\gamma_v = 3\alpha = 3 \times 1.0 \times 10^{-5} = 0.3 \times 10^{-4}\text{ K}^{-1}$ is the volume expansivity of the glass flask. Therefore, $\gamma_a = 5.0 \times 10^{-4} - 0.3 \times 10^{-4} = 4.7 \times 10^{-4}\text{ K}^{-1}$. The overflow volume is $\Delta V_a = V_0 \times \gamma_a \times \Delta T = 800 \times (4.7 \times 10^{-4}) \times 50 = 18.8\text{ cm}^3$.

Adım Adım Çözüm

1

Calculate the cubic expansivity of the glass flask

$\gamma_v = 3.0 \times 10^{-5}\text{ K}^{-1} = 0.3 \times 10^{-4}\text{ K}^{-1}$

The volume expansivity of a solid container is three times its linear expansivity ($\gamma_v = 3\alpha$).

2

Determine the apparent volume expansivity of the liquid

$\gamma_a = 4.7 \times 10^{-4}\text{ K}^{-1}$

Apparent expansivity is equal to real expansivity minus the vessel cubic expansivity ($\gamma_a = \gamma_r - \gamma_v$).

3

Calculate the change in temperature

$\Delta T = 50\text{ K}$

The temperature increases from

10^\circ\text{C}

to

60^\circ\text{C}

.

4

Calculate the overflow volume

$\Delta V_a = 18.8\text{ cm}^3$

The overflow volume is the apparent expansion of the liquid given by $\Delta V_a = V_0 \gamma_a \Delta T$.

Anahtar Kavram

Real and apparent cubic expansivity of liquids

Soru 166Soru →

A glass relative density bottle with linear expansivity $\alpha = 8.0 \times 10^{-6}\text{ K}^{-1}$ is completely filled with $204\text{ g}$ of a liquid at $15^\circ\text{C}$ . When heated to $65^\circ\text{C}$ , $4\text{ g}$ of the liquid overflows. What is the real cubic expansivity of the liquid?

4.24 \times 10^{-4}\text{ K}^{-1}

4.08 \times 10^{-4}\text{ K}^{-1}

4.00 \times 10^{-4}\text{ K}^{-1}

3.76 \times 10^{-4}\text{ K}^{-1}

Cevabı ve açıklamayı göster

Cevap: $4.24 \times 10^{-4}\text{ K}^{-1}$

Cevap

The real cubic expansivity of the liquid is

4.24 \times 10^{-4}\text{ K}^{-1}

.

The real cubic expansivity of a liquid accounts for both the apparent expansion observed as overflow and the expansion of the container itself. Calculating apparent cubic expansivity yields

\gamma_a = \frac{4\text{ g}}{200\text{ g} \times 50\text{ K}} = 4.0 \times 10^{-4}\text{ K}^{-1}

. Combining this with the glass bottle's cubic expansivity

\gamma_v = 3 \times 8.0 \times 10^{-6} = 0.24 \times 10^{-4}\text{ K}^{-1}

gives

\gamma_r = 4.24 \times 10^{-4}\text{ K}^{-1}

.

Adım Adım Çözüm

1

Calculate the mass of liquid remaining in the bottle at the higher temperature.

m_{\text{remaining}} = 204\text{ g} - 4\text{ g} = 200\text{ g}

The apparent expansivity formula using the mass method requires the remaining mass of liquid that occupies the bottle volume at the elevated temperature.

2

Calculate the apparent cubic expansivity (

\gamma_a

) of the liquid.

\gamma_a = \frac{\text{mass expelled}}{m_{\text{remaining}} \times \Delta T} = \frac{4}{200 \times (65 - 15)} = \frac{4}{10000} = 4.0 \times 10^{-4}\text{ K}^{-1}

Apparent expansivity is defined as the mass of liquid expelled divided by the product of remaining mass and temperature rise.

3

Determine the cubic expansivity of the glass vessel (

\gamma_v

).

\gamma_v = 3\alpha = 3 \times (8.0 \times 10^{-6}\text{ K}^{-1}) = 2.4 \times 10^{-5}\text{ K}^{-1} = 0.24 \times 10^{-4}\text{ K}^{-1}

The cubic expansivity of an isotropic solid container is three times its linear expansivity.

4

Calculate the real cubic expansivity of the liquid (

\gamma_r

).

\gamma_r = \gamma_a + \gamma_v = 4.0 \times 10^{-4}\text{ K}^{-1} + 0.24 \times 10^{-4}\text{ K}^{-1} = 4.24 \times 10^{-4}\text{ K}^{-1}

The real volume expansivity of a liquid equals the sum of its apparent expansivity and the cubic expansivity of its container.

Anahtar Kavram

Relationship between real cubic expansivity, apparent cubic expansivity, and vessel cubic expansivity (

\gamma_r = \gamma_a + \gamma_v

).

Tahmini Süre:1m 30s

Soru 167Soru →

A sample of liquid has an initial volume of $500\text{ cm}^3$ inside a container at $20^\circ\text{C}$ . Upon heating the system to $70^\circ\text{C}$ , the liquid's apparent volume expansion is measured to be $10\text{ cm}^3$ . If the linear expansivity of the container material is $1.0 \times 10^{-5}\text{ K}^{-1}$ , what is the real cubic expansivity of the liquid?

4.3 \times 10^{-4}\text{ K}^{-1}

4.1 \times 10^{-4}\text{ K}^{-1}

4.0 \times 10^{-4}\text{ K}^{-1}

3.7 \times 10^{-4}\text{ K}^{-1}

Cevabı ve açıklamayı göster

Cevap: $4.3 \times 10^{-4}\text{ K}^{-1}$

Cevap

The real cubic expansivity of the liquid is

4.3 \times 10^{-4}\text{ K}^{-1}

.

The real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubic expansivity of its vessel (

\gamma_r = \gamma_a + \gamma_v

). Calculating apparent cubic expansivity gives

\frac{10}{500 \times 50} = 4.0 \times 10^{-4}\text{ K}^{-1}

, and the container's cubic expansivity is

3 \times (1.0 \times 10^{-5}) = 0.3 \times 10^{-4}\text{ K}^{-1}

. Adding these yields

4.3 \times 10^{-4}\text{ K}^{-1}

.

Adım Adım Çözüm

1

Calculate the temperature change (

\Delta T

) and the apparent cubic expansivity (

\gamma_a

) of the liquid.

\Delta T = 70^\circ\text{C} - 20^\circ\text{C} = 50\text{ K}

, so

\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = \frac{10}{500 \times 50} = 4.0 \times 10^{-4}\text{ K}^{-1}

.

Apparent expansivity relates apparent volume increase to initial volume and temperature change.

2

Convert the container's linear expansivity (

\alpha_v

) to its cubic expansivity (

\gamma_v

).

\gamma_v = 3\alpha_v = 3 \times (1.0 \times 10^{-5}\text{ K}^{-1}) = 3.0 \times 10^{-5}\text{ K}^{-1} = 0.3 \times 10^{-4}\text{ K}^{-1}

.

Cubic expansivity of an isotropic solid container is three times its linear expansivity.

3

Calculate the real cubic expansivity of the liquid (

\gamma_r

) using the relation

\gamma_r = \gamma_a + \gamma_v

.

\gamma_r = 4.0 \times 10^{-4}\text{ K}^{-1} + 0.3 \times 10^{-4}\text{ K}^{-1} = 4.3 \times 10^{-4}\text{ K}^{-1}

.

Real expansion of a liquid accounts for both its observed (apparent) expansion and the expansion of the containing vessel.

Anahtar Kavram

Relationship between real expansivity, apparent expansivity of liquids, and container expansivity

Tahmini Süre:1m 30s

Soru 168Soru →

A flexible gas storage container at a research laboratory holds $0.30\text{ m}^3$ of helium gas at an initial temperature of $27^\circ\text{C}$ . If the gas is heated at constant pressure until its temperature reaches $127^\circ\text{C}$ , what is the final volume occupied by the gas?

0.40\text{ m}^3

1.41\text{ m}^3

0.23\text{ m}^3

0.10\text{ m}^3

Cevabı ve açıklamayı göster

Cevap: $0.40\text{ m}^3$

Cevap

The final volume of the gas is

0.40\text{ m}^3

.

According to Charles's law, at constant pressure, the volume of a gas is directly proportional to its absolute temperature (

V \propto T

). Converting

27^\circ\text{C}

to

300\text{ K}

and

127^\circ\text{C}

to

400\text{ K}

, the volume expands by a ratio of

\frac{400}{300} = \frac{4}{3}

. Multiplying the initial volume

0.30\text{ m}^3

by

\frac{4}{3}

gives

0.40\text{ m}^3

.

Adım Adım Çözüm

1

Convert initial and final temperatures from Celsius to Kelvin

T_1 = 27 + 273 = 300\text{ K}

,

T_2 = 127 + 273 = 400\text{ K}

Gas laws require absolute temperatures in Kelvin to maintain proportional relationships.

2

Apply Charles's Law for constant pressure processes

\frac{V_1}{T_1} = \frac{V_2}{T_2} \implies V_2 = V_1 \times \frac{T_2}{T_1}

At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.

3

Substitute the known values and calculate the final volume

V_2 = 0.30\text{ m}^3 \times \frac{400\text{ K}}{300\text{ K}} = 0.40\text{ m}^3

Performing arithmetic yields the expanded volume.

Anahtar Kavram

Charles's Law

Soru 169Soru →

A high-altitude research chamber of fixed volume contains nitrogen gas at an initial pressure of $1.00 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The gas is heated until its pressure rises to $2.20 \times 10^5\text{ Pa}$ . What is the final temperature of the gas in degrees Celsius?

Cevabı ve açıklamayı göster

Cevap: 387

Cevap

387

By Gay-Lussac's Law at constant volume, pressure is directly proportional to absolute temperature. Converting

27^\circ\text{C}

to

300\text{ K}

, the final temperature is

300 \times \frac{2.20 \times 10^5}{1.00 \times 10^5} = 660\text{ K}

, which equals

387^\circ\text{C}

.

Adım Adım Çözüm

1

Convert initial temperature to Kelvin

T_1 = 300 K

Gas laws require absolute temperature in Kelvin.

2

Apply Pressure Law (P1 / T1 = P2 / T2)

T_2 = 660 K

Since volume is fixed, pressure is directly proportional to absolute temperature.

3

Convert absolute temperature back to Celsius

t_2 = 387 °C

The question asks for the temperature in degrees Celsius.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law)

Soru 170Soru →

At a temperature of $27^\circ\text{C}$ , the root-mean-square (r.m.s.) speed of the molecules of an ideal gas is $300\text{ m/s}$ . What is the temperature of the gas, in degrees Celsius, when the r.m.s. speed of its molecules increases to $600\text{ m/s}$ ?

Cevabı ve açıklamayı göster

Cevap: 927

Cevap

The temperature of the gas when the r.m.s. speed reaches

600\text{ m/s}

is

927^\circ\text{C}

.

According to kinetic theory, the root-mean-square speed of gas molecules is directly proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

). First convert the initial temperature to Kelvin:

27^\circ\text{C} + 273 = 300\text{ K}

. Since the speed doubles from

300\text{ m/s}

to

600\text{ m/s}

, the ratio of speeds is

2

, which means the absolute temperature ratio is

2^2 = 4

. Thus, the new absolute temperature is

4 \times 300\text{ K} = 1200\text{ K}

. Converting back to Celsius gives

1200 - 273 = 927^\circ\text{C}

.

Adım Adım Çözüm

1

Convert the initial temperature from Celsius to Kelvin

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

Gas kinetic equations require absolute temperature in Kelvin.

2

Apply the proportional relationship between r.m.s. speed and absolute temperature

\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}

In the kinetic theory of gases, root-mean-square speed is directly proportional to the square root of absolute temperature.

3

Calculate the final absolute temperature

T_2

T_2 = 1200\text{ K}

Doubling the r.m.s. speed requires quadrupling the absolute temperature (

2^2 \times 300\text{ K} = 1200\text{ K}

).

4

Convert the calculated absolute temperature back to degrees Celsius

\theta_2 = 1200 - 273 = 927^\circ\text{C}

Subtract 273 from the Kelvin temperature to find the value in degrees Celsius.

Anahtar Kavram

Proportionality between root-mean-square speed and absolute temperature in kinetic theory of gases

Thermal Physics

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