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Bumek [7]
3 years ago
11

A certain gas is present in a 10.0 LL cylinder at 4.0 atmatm pressure. If the pressure is increased to 8.0 atmatm the volume of

the gas decreases to 5.0 LL . Find the two constants kiki, the initial value of kk, and kfkf, the final value of kk, to verify whether the gas obeys Boyle’s law by entering the numerical values for kiki and kfkf in the space provided.
Chemistry
1 answer:
ICE Princess25 [194]3 years ago
6 0

Answer:

The gas obeys Boyle’s law and the value of k_i\&k_f both are equal to 40.0 atm L.

Explanation:

Initial volume of the gas = V_1=10.0 L

Initial pressure of the gas = P_1=4.0 atm

Final volume of the gas = V_2=5.0 L

Final pressure of the gas = P_2=8.0 atm

This law states that pressure is inversely proportional to the volume of the gas at constant temperature.  

PV=k

The equation given by this law is:

P_1V_1=P_2V_2

P_1\propto \frac{1}{V_1}

P_1V_1=k_i

k_i=4.0 atm\times 10.0 L = 40.0 atm L

P_2\propto \frac{1}{V_2}

P_V_2=k_f

k_f=8.0 atm\times 5.0 L = 40.0 atm L

k_i=k_f=40.0 atm L

The gas in the cylinder is obeying Boyle's law.

The gas obeys Boyle’s law and the value of k_i\&k_f both are equal to 40.0 atm L.

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