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Arturiano [62]
3 years ago
9

0.0625 mol of an ideal gas is contained within a 1.0000 Lvolume container. Its pressure is 142,868 Pa and temperature is X K . W

hat is X?
(HINT: |X| is near an order of magnitude of 102 K ).
Chemistry
1 answer:
aliya0001 [1]3 years ago
8 0

<u>Answer:</u> The temperature of the ideal gas is 2.75\times 10^2K

<u>Explanation:</u>

To calculate the temperature, we use the equation given by ideal gas equation:

PV=nRT

where,

P = Pressure of the gas = 142,868 Pa = 142.868 kPa    (Conversion factor: 1 kPa = 1000 Pa)  

V = Volume of gas = 1.0000 L

n = number of moles of ideal gas = 0.0625 moles

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

T = temperature of the gas = ?

Putting values in above equation, we get:

142.868kPa\times 1.0000=0.0625mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times T\\\\T=275K=2.75\times 10^2K

Hence, the temperature of the ideal gas is 2.75\times 10^2K

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Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
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Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0
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What happens to ice as it warms up
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Ice starts to melt and turns into water as it melts
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