At a temperature of 320K, the gas in a cylinder has a volume of 40.0 liters. If the volume of the gas is decreased to 20.0 liters, what must the temperature be for the gas pressure to remain constant?A. 160 K
B. 273 K
C. 560 K
D. 140 K

Answers

Answer 1
Answer: Assuming the gas behaves ideally,
PV/T = constant. P will also be constant in this giving us:
V₁/T₁ = V₂/T₂
40/320 = 20/T₂
T₂ = 160 K
The answer is A.
Answer 2
Answer:

Answer:

The correct answer is option A.

Explanation:

Initial volume of the gas V_1= 40.0L

Initial temperature of gas T_1= 320 K

Final volume of the gas V_2= 20.0L

Final temperature of the gas = T_2

Applying Charles' Law:

(V_1)/(T_1)=(V_2)/(T_2)

T_2=(V_2* T_1)/(V_1)=(20.0L* 320 K)/(40.0L)=160K

The temperature of the gas when volume of the gas is 20.0 L is 160 K.Hence, the correct answer is option A.


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Answers

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A current of 120 mA flows past a point in a circuit for 25 s. How much charge passes the point in this time?
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Answers

a) I = 120 mA = 120 x 10^(-3) A = 0.12
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Answers

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Explanation:

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Answers

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Answers

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