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Brums [2.3K]
3 years ago
14

A gas is in a sealed container.Part ABy what factor does the gas temperature change if the volume is doubled and the pressure is

tripled?Part BBy what factor does the gas temperature change if the volume is halved and the pressure is tripled?
Physics
1 answer:
Goryan [66]3 years ago
4 0

A) The temperature increases by a factor 6

We can use the ideal gas equation:

pV=nRT

where

p is the pressure

V is the volume

n is the number of moles

R is the gas constant

T is the temperature

We can also rewrite it as

\frac{pV}{T}=nR

The gas is in a sealed container - this means the amount of gas is fixed, so n is constant. Since R is constant too, the term on the right in the equation is constant. So we can rewrite the equation as:

\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}

where in this problem we have:

V_2 = 2V_1 (the volume is doubled)

p_2 = 3 p_1 (the pressure is tripled)

re-arranging the equation, we find the change in temperature:

\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(2 V_1)}{p_1 V_1}=6

so, the temperature increases by a factor 6.

B) The temperature increases by a factor 1.5.

We can use again the same equation:

\frac{p_1V_1}{T_1}=\frac{p_2 V_2}{T_2}

Where in this case:

V_2 = \frac{V_1}{2} (the volume is halved)

p_2 = 3 p_1 (the pressure is tripled)

So, we can find the change in temperature:

\frac{T_2}{T_1}=\frac{p_2 V_2}{p_1 V_1}=\frac{(3 p_1)(\frac{V_1}{2})}{p_1 V_1}=\frac{3}{2}=1.5

So, the temperature increases by 1.5 times.

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3 years ago
A block of ice (m = 9 kg) at a temperature of T1 = 0 degrees C is placed out in the sun until it melts, and the temperature of t
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Answer:

a) An expression for the amount of energy, E_m, needed to melt the ice into water.

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b) An expression for the total amount of energy, E_tot, to melt the ice and then bring the water to T2

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c) 3,646,458 J = 3646.46 kJ

Explanation:

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(E_m) = (m × Lf)

b) The Heat required to raise the temperature of a body from one temperature to another is given by the product of the mass of the body, its specific heat capacity and the temperature difference between the final point and the starting point.

(E_2) = mcΔT = mc (T2 - T1)

Total heat required to melt the ice at T1 = 0 and raise the temperature of the resulting water to T2 is then a sum of (E_m) + (E_2)

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c) What is the energy in Joules?

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c = Specific heat capacity of water = 4186 J/kg.K

T2 = final temperature of the water = 17°C

T1 = Initial temperature of the water = 0°C

Note that the units of temperature difference is the same for K and °C

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Q = 3,006,000 + 640,458 = 3,646,458 J = 3646.46 kJ

Hope this Helps!!!

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