Question

a 500 ml container has 15×10¹² molecules of an ideal gas .how many molecules will occupy 100 ml at same temperature and pressure
(a) 3 × 10¹²
(b) 75 × 10 ¹²
(c) 7.5 × 10 ¹²​

Answer 1

Answer:

The correct option is (a) $3\times 10^{12}$.

Explanation:

Given:

For the volume, $\rm V_1=500\ ml$, number of molecules, $\rm N_1 = 15\times 10^{12}.$

Let there are $\rm N_2$ number of molecules in the volume $\rm V_2=100\ ml.$

Since, the gas is given to be ideal gas, therefore it follows the ideal gas law, which is,

$\rm PV=nRT=NkT.$

where,

• P = pressure of the gas.

• V = volume of the gas.

• T = absolute temperature of the gas.

• R = Universal gas constant.

• k = Boltzmann constant.

• n = number of moles of the gas.

• N = number of molecules of the gas.

We are given that the temperature and pressure of the gas is same.

Applying ideal gas law for the volume $\rm V_1$,

$\rm PV_1=N_1kT.$

Applying ideal gas law for the volume $\rm V_2$,

$\rm PV_2=N_2kT.$

Dividing both these equations, we get,

$\rm \dfrac{PV_1}{PV_2}=\dfrac{N_1kT}{N_2kT}\\ \dfrac{V_1}{V_2}=\dfrac{N_1}{N_2}\\N_2 = \dfrac{V_2}{V_1}\times N_1 = \dfrac{100\ ml}{500\ ml}\times15\times 10^{12} =3\times 10^{12}\ molecules.$

Thus, correct option is (a) $3\times 10^{12}$.

Answer 2

Answer:

See the attachment

Good luck

Explanation: