Consider a fixed volume and amount of an ideal gas at 10 °C.
What temperature would it have to be changed to in order for the pressure to be doubled?
A. 5°C
B. 20 °C
C. 566 °C
D. 293 °C
E. 278 °C
This question will be dealing with the ideal gas law equations, we can be summed up through:
PV=nRT
We are told that we have an ideal gas at a temperature of 10°C and we want to know what temperature increase would cause the doubling in pressure.
Want:
T_2
n,R, and V are constants
Leaves us with an equation of:
\frac{P_1}{T_1} = \frac{P_2}{T_2}
We want to find T_2 when we have double the pressure, so when P_2 = 2P, where P_1 = P
We also know T_1 = 10°C, which can be converted to K.
10°C = (273 + 10)K T_1 = 283K
Now sub all this into our equation:
\frac{P}{283K} = \frac{2P}{T_2}
P can be anything, it does not matter, what is important is the relationship between P_1 and P_2, which is that P_2 is double. In order to solve this easier, we can assign P a random value and then solve. P=1 is the simplest way.
\frac{1}{283K} = \frac{2(1)}{T_2}
\frac{T_2}{283K} = {2}
{T_2} = (2)(283K)
{T_2} = 566K
Now we just need to convert to °C.
$556K - 273 = 293°C
\fcolorbox{red}{grey!30}{Therefore the answer is D, $293K$.}
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