Find the number of molecules in 0.25\ m^3 of the gas at a pressure 2.0 × 10^2\ kPa and a temperature 300\ K . [R = 8.314\ J K^ -1 mol^ -1 ]

Correct answer: A. 1.2044 × 10^ 25 Given: Pressure P = 2.0 × 10^2\ kPa = 2.0 × 10^5\ Pa Volume V = 0.25\ m^3 Temperature T = 300\ K Universal gas constant R = 8.314\ J K^ -1 mol^ -1 Using the ideal gas equation, the number of moles n is: PV = nRT n = (PV)/(RT) = (2.0 × 10^5 × 0.25)/(8.314 × 300) n = (50000)/(2494.2) 20\ mol The total number of molecules N is given by: N = n × N_A Substituting Avogadro's number N_A = 6.022 × 10^ 23 \ mol^ -1 : N = 20 × 6.022 × 10^ 23 = 1.2044 × 10^ 25 Answer: 1.2044 × 10^ 25

MHT CET 202621 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 21 April
Shift: Evening Shift
Subject: Chemistry
Chapter: States of Matter

MHT CET 2026 · 21 April · Evening Shift · Chemistry · States of Matter

Find the number of molecules in 0.25\ m^3 of the gas at a pressure...

Find the number of molecules in

0.25\ \text{m}^3

of the gas at a pressure

2.0 \times 10^2\ \text{kPa}

and a temperature

300\ \text{K}

[R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}]

Options

A $1.2044 \times 10^{25}$
B $2.4088 \times 10^{25}$
C $3.6077 \times 10^{25}$
D $4.8099 \times 10^{25}$

Correct answer

A. $1.2044 \times 10^{25}$

Step-by-step solution

Given:
Pressure

P = 2.0 \times 10^2\ \text{kPa} = 2.0 \times 10^5\ \text{Pa}

Volume

V = 0.25\ \text{m}^3

Temperature

T = 300\ \text{K}

Universal gas constant

R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}

Using the ideal gas equation, the number of moles

n

is:

PV = nRT

n = \dfrac{PV}{RT} = \dfrac{2.0 \times 10^5 \times 0.25}{8.314 \times 300}

n = \dfrac{50000}{2494.2} \approx 20\ \text{mol}

The total number of molecules

N

is given by:

N = n \times N_A

Substituting Avogadro's number

N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}

N = 20 \times 6.022 \times 10^{23} = 1.2044 \times 10^{25}

Answer:

1.2044 \times 10^{25}

Practice States of Matter on Quantrex Academy →

More from States of Matter