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anonymous
 4 years ago
A fixed amount of ideal gas is held in a rigid container that expands negligibly when heated. At 20°C the gas pressure is P. If we add enough heat to increase the temperature from 20°C to 40°C, the pressure will be:
A. equal to 2P
B. less than 2P
C. impossible to determine since we do not know the number of moles of gas in the container
D. greater than 2P
or
E. impossible to determine since we do not know the volume of gas in the container.
anonymous
 4 years ago
A fixed amount of ideal gas is held in a rigid container that expands negligibly when heated. At 20°C the gas pressure is P. If we add enough heat to increase the temperature from 20°C to 40°C, the pressure will be: A. equal to 2P B. less than 2P C. impossible to determine since we do not know the number of moles of gas in the container D. greater than 2P or E. impossible to determine since we do not know the volume of gas in the container.

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1By the ideal gas law, PV/T = constant. Use this fact to answer your question.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, since v isnt given, we can just ignore it. So a comparison like this would be appropriate: \[P_{1}/2 = P_{2}/4\rightarrow 2P _{1}=P _{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So the answer is then A.2P! right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not exactly, be careful about T! This has to be the absolute temperature, i.e. the temperature in Kelvin.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well i just meant 20c and 40c

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1In the ideal gas law, we have to use temperature in Kelvin. Otherwise we would very quickly have problems. For example, given that P/T is a constant, suppose T was in Celcius and we wanted to know the pressure for a balloon of hydrogen gas going from 1 atm at 20 C to 10 C. Then we'd have \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = \frac{P_1}{T_1} T_2 = \frac{1}{20}(10) = 1/2 \] a negative pressure! Which is clearly nonsense. Hence you need to convert the temperatures 20 and 40 C into an absolute temperature scale, Kelvin. E.g. 20 C = 273 + 20 K = 293 K

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But would my above proportion hold true?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No, because \[T_1 = (20 + 273) K = 293 K\]\[T_2 = (40 + 273) K = 313 K\] and therefore \[\frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{313 K}{293 K} < 2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So how does that use PV/T

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1PV/T is a constant. Hence if V is also a constant, then P/T is a constant also.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So if we can not use the relation p1/293k = p2/313k, Im not sure how to go about with a solution

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1...and \[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \implies \frac{P_2}{P_1} = \frac{T_2}{T_1} \] It is this second form that chris has used in his solution. Think about the algebra here a bit more before being so reflexive.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We are trying to find P2, and in our answers we can use P1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So why not muliply P1 to the other side?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0P2 = 313P2/293 > P2 = 1.068ishP2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is the same answer I could get with my original proportion of P1/293 = P2/313, no?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Mess up the P2 a bit in a few posts above.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just needed to convert

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, I get what you guys were saying, haha. thanks!
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