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3 years ago

14

A tank is filled with an ideal gas at 400 K and pressure of 1.00 atm.

1 answer:

bekas [8.4K]3 years ago

6 0

To find the temperature it is necessary to use the expression and concepts related to the ideal gas law.

Mathematically it can be defined as

$PV=nRT$

Where

P = Pressure

V = Volume

n = Number of moles

R = Gas constant

T = Temperature

When the number of moles and volume is constant then the expression can be written as

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Or in practical terms for this exercise depending on the final temperature:

$T_2 = \frac{P_2T_1}{P_1}$

Our values are given as

$T_1 = 400K\\P_1 = 1atm\\P_2 = 2atm$

Replacing

$T_2 = \frac{(2)(400)}{1}\\T_2 = 800K$

Therefore the final temperature of the gas is 800K

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3 years ago

Read 2 more answers

Ilia_Sergeevich [38]

Answer:

The value is $C = 30729\ c$

Explanation:

From the question we are told that

The power rating of the stove is $P = 4.4 KW = 4.4 *10^{3} \ W$

The duration of its use everyday is $t_1 = 70 \ minutes = 1.167 \ hours$

The rating of the light bulbs is $P_2' = 150 W$

The number is $n = 7$

The power rating of the total bulb is $P_2 = 7 * 150 = 1050 \ W$

The duration of its use everyday is $t_2 = 7 hours$

The power rating of miscellaneous appliance $P_3 = 1.8 \ KW = 1.8 *10^{3} \ W$

The duration of its use everyday is $t_3 = 1 hour$

The power rating of hot water $P = 4 KW = 4 *10^{3} \ W$

The duration of its use everyday is $t_4 = 120 \ minutes = 2 hours$

Generally the total electrical energy used in 1 month is mathematically represented as

$E = P_1 * t_1 * 30 + P_2 * t_2 * 30 + P_3 * t_3 * 30+ P_4 * t_4 * 30$

=> $E = 4.4*10^{3} * 1.167 * 30 + 1050 * 7 * 30 + 1.8 *1 * 30+ 4*10^{3} * 2 * 30$

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3 years ago

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3 years ago

what is the maximum speed of a 34-g object bouncing on a spring (k=78.1 N/m) with an amplitude of 3.5-cm?

Vika [28.1K]

Answer:

The maximum speed of the mass is 1.67 m/s.

Explanation:

We have,

Mass of object is 34 g or 0.034 kg

Spring constant of the spring is 78.1 N/m

Amplitude attained by the object is 3.5 cm or 0.035 m

It is required to find the maximum speed of the object in this spring mass system. The maximum speed is given by :

$v=A\omega$

$\omega=\sqrt{\dfrac{k}{m}}$

$v=A\sqrt{\dfrac{k}{m}}$

Plugging all the values in above formula,

$v=0.035\times \sqrt{\dfrac{78.1}{0.034}}\\\\v=1.67\ m/s$

So, the maximum speed of the mass is 1.67 m/s.

4 0

3 years ago