Question

A certain gas is present in a 13.0 l cylinder at 4.0 atm pressure. if the pressure is increased to 8.0 atm , the volume of the gas decreases to 6.5 l . find the two constants ki, the initial value of k, and kf, the final value of k, to verify whether the gas obeys boyle's law. express your answers to two significant figures separated by a comma.

Answer 1

Answer:

Yes, gas is obeying the Boyle's law.

Explanation:

Boyle's Law  states that 'pressure is inversely proportional to the volume of the gas at constant temperature and number of moles'.

$P\propto \frac{1}{V}$     (At constant temperature and number of moles)

$P_1V_2 = constant=k=P_2V_2$

Initial volume of the gas = $V_1=13.0L$

Initial pressure of the gas = $P_1=4.0 atm$

Initial value of k :

$k_i=P_1V_1=4 atm \times 13.0 L =52 atm l$

Final volume of the gas = $V_2=6.5 L$

Final pressure of the gas = $P_2=8.0 atm$

Final value of k :

$k_f=P_2V_2=8.0 atm \times 6.5 L =52 atm l$

$k_i=k_f=52 atm L$

As we can see that initial value of k is equal to the final value of k which means that gas is obeying the Boyle's law.

Answer 2

Boyle's Law is k = PV so Initial k = 13.0 L x 4.0 atm = 52 L atm Final kf = 6.5 L x 8 atm = 52 L atm The gas obeys Boyle's Law The answer with two significant figures separated by a comma is k = 52, kf = 52.