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3 years ago

4. In a object filled with a type of gas, which of the following actions would decrease the pressure

Physics

1 answer:

ExtremeBDS [4]3 years ago

7 0

Answer:

Decrease the number of gas particles, increase the object's area, and reduce the temperature of the gas

Explanation:

TEMPERATURE:

Decreasing the temperature will slow down the molecules. Hence, less no. of collisions will take place between walls of object and molecules. This will result in decrease of pressure.

Therefore, the pressure of a gas can be decreased by increasing its temperature.

NUMBER OF GAS PARTICLES:

Decreasing the number of particles will result in less no. of collisions, hence decreasing the pressure.

Therefore, the pressure of a gas can be decreased by decreasing its no. of molecules or no. of particles.

AREA OF OBJECT:

The pressure is given by the formula:

$P = \frac{F}{A}\\\\For constant Force (F):\\P\ \alpha\ \frac{1}{A}$

where,

A = Area of Object

Therefore, the pressure of a gas can be decreased by increasing area of object.

So, the correct option is:

Decrease the number of gas particles, increase the object's area, and reduce the temperature of the gas

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3 years ago

A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.