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Question 1521Question →

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?

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Answer: 387

Answer

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}

.

Step-by-Step Solution

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.

Key Concept

Pressure Law (Gay-Lussac's Law)

Question 1522Question →

A machine gun fires bullets, each of mass $0.020\text{ kg}$ , at a speed of $400\text{ m s}^{-1}$ . If the gun fires $5\text{ bullets per second}$ , what is the magnitude of the average recoil force exerted on the gun in newtons?

Show answer & explanation

Answer: 40

Answer

The magnitude of the average recoil force exerted on the gun is

40\text{ N}

.

According to Newton's second law of motion, force is equal to the rate of change of momentum (

F = \frac{\Delta p}{\Delta t}

). The total mass of ammunition leaving the gun per second is

5 \times 0.020\text{ kg} = 0.10\text{ kg s}^{-1}

. Multiplying this mass rate by the velocity (

400\text{ m s}^{-1}

) gives a rate of momentum change of

40\text{ N}

, which corresponds directly to the average recoil force.

Step-by-Step Solution

1

Calculate the mass of bullets fired per unit time (mass flow rate).

\frac{\Delta m}{\Delta t} = 5 \text{ bullets/s} \times 0.020 \text{ kg/bullet} = 0.10 \text{ kg s}^{-1}

.

Force is defined as the rate of change of linear momentum, which requires knowing the total mass delivered per second.

2

Apply Newton's second law in terms of momentum change per second.

F = \frac{\Delta p}{\Delta t} = \left(\frac{\Delta m}{\Delta t}\right) v = 0.10 \text{ kg s}^{-1} \times 400 \text{ m s}^{-1} = 40 \text{ N}

.

The rate of momentum change of the bullets equals the magnitude of the force exerted on them, which by Newton's third law equals the recoil force on the gun.

Key Concept

Newton's Second Law of Motion and Rate of Change of Linear Momentum

Question 1523Question →

A cell with an electromotive force (e.m.f.) of $12.0\text{ V}$ and an internal resistance of $1.5\ \Omega$ is connected in series to an external resistor of $8.5\ \Omega$ . What is the terminal potential difference across the cell?

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Answer: 10.2

Answer

The terminal potential difference across the cell is

10.2\text{ V}

.

The terminal potential difference

V

across a real cell supplying current is less than its electromotive force

E

due to the internal voltage drop

Ir

. By first determining the circuit current

I = \frac{E}{R + r} = \frac{12.0}{8.5 + 1.5} = 1.2\text{ A}

, the terminal potential difference is calculated as

V = I \times R = 1.2 \times 8.5 = 10.2\text{ V}

(or equivalently

V = E - I r = 12.0 - (1.2 \times 1.5) = 10.2\text{ V}

).

Step-by-Step Solution

1

Calculate the total resistance of the circuit

R_{\text{total}} = 10.0\ \Omega

The external load resistor and the cell's internal resistance are in series.

2

Determine the total current drawn from the cell

I = 1.2\text{ A}

Using Ohm's law applied to the entire circuit,

I = \frac{E}{R + r}

.

3

Calculate the potential drop across the external load resistor

V = 10.2\text{ V}

Terminal voltage equals potential difference across external load (

V = I R

) or

V = E - I r

.

Key Concept

Terminal Potential Difference and Internal Resistance

Question 1524Question →

An electric immersion heater rated at $1.5\,\text{kW}$ is operated on a $240\,\text{V}$ mains supply for $14\,\text{minutes}$ . Calculate the total electrical energy consumed by the heater during this period, in megajoules ( $\text{MJ}$ ).

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Answer: 1.26

Answer

The total electrical energy consumed by the heater is

1.26\,\text{MJ}

.

Electrical energy consumed is obtained by multiplying electrical power by time (

E = P \times t

). Converting power to watts (

1.5\,\text{kW} = 1500\,\text{W}

) and time to seconds (

14\,\text{min} = 840\,\text{s}

) yields

E = 1500 \times 840 = 1,260,000\,\text{J} = 1.26\,\text{MJ}

.

Step-by-Step Solution

1

Convert power rating from kilowatts to watts

P = 1.5\,\text{kW} = 1500\,\text{W}

The standard SI unit of power for energy calculation in Joules is Watts.

2

Convert time duration from minutes to seconds

t = 14\,\text{minutes} \times 60\,\text{seconds/minute} = 840\,\text{seconds}

The standard SI unit of time in Joule calculations is seconds.

3

Calculate energy consumed in Joules using

E = P \times t

E = 1500\,\text{W} \times 840\,\text{s} = 1,260,000\,\text{J}

Electrical energy is the product of power in watts and time in seconds.

4

Convert energy from Joules to Megajoules

E = \frac{1,260,000}{10^6} = 1.26\,\text{MJ}

One Megajoule (

1\,\text{MJ}

) is equivalent to

10^6\,\text{Joules}

.

Key Concept

Electrical Energy Consumption

Question 1525Question →

At a meteorological station in Maiduguri, Nigeria, the maximum shade air temperature recorded during a 24-hour period was $39.4^\circ\text{C}$ , while the minimum temperature recorded was $21.8^\circ\text{C}$ . What is the diurnal temperature range in degrees Celsius for that day?

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Answer: 17.6

Answer

The diurnal temperature range for the day is

17.6^\circ\text{C}

.

The diurnal range of temperature is defined as the difference between the highest (maximum) and lowest (minimum) temperatures recorded within a 24-hour period. Subtracting the minimum temperature (

21.8^\circ\text{C}

) from the maximum temperature (

39.4^\circ\text{C}

) yields

17.6^\circ\text{C}

.

Step-by-Step Solution

1

Identify the recorded daily maximum and minimum temperatures.

Maximum temperature =

39.4^\circ\text{C}

, Minimum temperature =

21.8^\circ\text{C}

.

Diurnal temperature range requires the highest and lowest values recorded during a 24-hour cycle.

2

Apply the diurnal temperature range formula.

Diurnal Range = Maximum Temperature - Minimum Temperature

The diurnal range represents the arithmetic difference between the maximum and minimum temperatures of the day.

3

Subtract the minimum temperature from the maximum temperature.

39.4^\circ\text{C} - 21.8^\circ\text{C} = 17.6^\circ\text{C}

Executing the subtraction gives the exact daily temperature variation.

Key Concept

Diurnal Temperature Range Calculation

Estimated Time:1m 0s

Question 1526Question →

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 $600\text{ m/s}$ ?

Show answer & explanation

Answer: 927

Answer

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}

.

Step-by-Step Solution

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.

Key Concept

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

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