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inysia [295]
3 years ago
14

A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6 m^3. How much work is done by the gas if it expands at a co

nstant pressure to twice its initial volume.
Physics
1 answer:
Ivenika [448]3 years ago
5 0
In thermodynamics, work of a system at constant pressure conditions is equal to the product of the pressure and the change in volume. It is expressed as follows:

W = P(V2 - V1)
W = 1.3x10^5 (2x6 - 6 )
<span>W = 780000 J
</span>
Hope this answers the question. Have a nice day.
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Which of the following statements is false? Group of answer choices Light that is high in energy has a high frequency. The atomi
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The last statement is false.

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Photons (Electromagnetic radiation) are released when electrons drop from a higher energy lever to a lower energy level. Therefore the opposite insinuated by the last statement is wrong.

4 0
3 years ago
The time T in seconds for a pendulum of length L feet to make one swing is given by Upper T=2\pi \sqrt((L)/(36)). How long is a
Andreyy89

Answer:

3.6ft

Explanation:

Using= 2*π*sqrt(L/32)

To solve for L, first move 2*n over:

T/(2*π) = sqrt(L/32)

Next,eliminate the square root by squaring both sides

(T/(2*π))2 = L/32

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T2/(4π2) = L/32

Lastly, multiply both sides by 32 to yield:

32T2/(4π2) = L

and simplify:

8T²/π²= L

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5 0
3 years ago
Las condiciones iniciales de un gas son 3000 cm3
slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

PV=nRT

V: volume = 3000cm^3 = 3L

P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

T: temperature = 27°C = 300.15K

For the given values you firs calculate the number n of moles:

n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C

hence, T'=92.70°C

8 0
3 years ago
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