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Natasha2012 [34]
2 years ago
9

A cylinder contains 3.0 L of oxygen at 310 K and 2.5 atm. The gas is heated, causing a piston in the cylinder to move outward. T

he heating causes the temperature to rise to 610 K and the volume of the cylinder to increase to 9.5 L. What is the gas pressure? P= _____atm
Physics
1 answer:
Alex_Xolod [135]2 years ago
5 0

Answer:

The gas pressure is: 1.55 atm.

Explanation:

We need to use the equation that relate the variables given at the exercise (pressure, temperature and volume) from the ideal gas law formula, when the mass is constant we can reduce the expretion PV=nRT to \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2} } solving to P2 we get:\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}=P_{2} replace the values P_{2}=\frac{2.5*3*610}{9.5*310} =1.55(atm).

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bulgar [2K]

Answer:

\frac{1}{2}

Explanation:

\frac{5}{6} -\frac{1}{3} = \frac{15}{18} -\frac{6}{18}=\frac{9}{18} =\frac{1}{2}

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2 years ago
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Biological weathering

Explanation:

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2 years ago
Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of
mart [117]

Answer:

The value is  B =  3.33 *10^{-6} \  T

Explanation:

From the question we are told that

  The distance of separation is  d = 0.6 \  m

  The current on the one wire is I_1 =  9 \  A

  The current on the second wire is I_2 =  4 \ A

Generally the magnitude of the field exerted between the current carrying wire is

        B  =  B_1 - B_2

Here B_1 is the magnetic field due to the first wire which is mathematically represented as

         B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}

Here d_1 is the distance to the half way point of the separation and the value is  

    d_1 =  0.3 \  m

B_2 is the magnetic field due to the first wire which is mathematically represented as

         B_2  = \frac{\mu_o * I_2 }{2 \pi * d_2}

Here d_2 is the distance to the half way point of the separation and the value is  

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This means that d_1 = d_2 = a =  0.3

So

     B =  \frac{\mu_o * I_1 }{2 \pi * d_1}  -  \frac{\mu_o * I_2 }{2 \pi * d_2}

=>  B =  \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }

=>  B =  \frac{  4\pi * 10^{-7}  * (9- 4)}{2 * 3.142  *0.3 }

=>  B =  3.33 *10^{-6} \  T

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2 years ago
A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat
KonstantinChe [14]

Answer:

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Explanation:

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s_w = Specific heat of water

L_i = Latent heat of ice

v = Velocity of ice

\Delta T = Change in temperature

Amount of heat required for steam

Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)

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Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6

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Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)

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Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m

The kinetic energy of the bullet will balance the heat

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Explanation:

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