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Answer:
the heat rate required to cool down the gas from 535°C until 215°C is -2.5 kW.
Explanation:
assuming ideal gas behaviour:
PV=nRT
therefore
P= 109 Kpa= 1.07575 atm
V= 67 m3/hr = 18.6111 L/s
T= 215 °C = 488 K
R = 0.082 atm L /mol K
n = PV/RT = 109 Kpa = 1.07575 atm * 18.611 L/s /(0.082 atm L/mol K * 488 K)
n= 0.5 mol/s
since the changes in kinetic and potencial energy are negligible, the heat required is equal to the enthalpy change of the gas:
Q= n* Δh = 0.5 mol/s * (- 5 kJ/mol) =2.5 kW
The volume increases when the balloon temperature increases.
<u>Explanation:</u>
-10 F is converted into Kelvin as 249 K.
0°C is nothing but 0+ 273 = 273 K
And the room temperature is 25°C which is converted into Kelvin as 273 + 25 = 298 K.
249 K is below room temperature.
As per the Charles' law volume and temperature are directly proportional to each other, when the pressure of the gas remains constant.
V ∝ T
As the balloon temperature increases, the volume also increases.
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