Asked by Anonymous
Very confused with this question. Amy help would be appreciated.
The ideal gas law states that P V = nRT where P is the pressure in atmospheres, V is the volume in litres, n is the number of moles, R = 0.082 L·atm/K·mol is the gas constant, and T is the temperature in Kelvins. Consider two moles of gas with 5 atmospheres of pressure and a volume of 15 L. Suppose the pressure is decreasing at 0.3 atm/min and the volume is increasing at 0.6 L/min. Given that the number of moles stays constant, what is the rate of change of the temperature with respect to time?
The ideal gas law states that P V = nRT where P is the pressure in atmospheres, V is the volume in litres, n is the number of moles, R = 0.082 L·atm/K·mol is the gas constant, and T is the temperature in Kelvins. Consider two moles of gas with 5 atmospheres of pressure and a volume of 15 L. Suppose the pressure is decreasing at 0.3 atm/min and the volume is increasing at 0.6 L/min. Given that the number of moles stays constant, what is the rate of change of the temperature with respect to time?
Answers
Answered by
Steve
Since moles is constant, nR is constant. SO,
V dP/dt + P dV/dt = nR dT/dt
You now have n = 2*6.02*10^23
P = 5
V = 15
dP/dt = -3
dV/dt = 0.6
So, plug in the numbers and solve for dT/dt
V dP/dt + P dV/dt = nR dT/dt
You now have n = 2*6.02*10^23
P = 5
V = 15
dP/dt = -3
dV/dt = 0.6
So, plug in the numbers and solve for dT/dt
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