Calculate the pressure in atm of 48 g oxygen gas in a 2\ dm^3 container at 27\ ^ ^ C . [ R = 0.0821\ atm dm^3\ ^ -1 \ mol^ -1 ]

Correct answer: A. 18.47\ atm Given mass of oxygen gas, w = 48\ g Molar mass of oxygen gas ( O_2 ), M = 32\ g mol^ -1 Number of moles, n = (w)/(M) = (48)/(32) = 1.5\ mol Volume of the container, V = 2\ dm^3 Temperature, T = 27\ ^ ^ C = 27 + 273 = 300\ K Using the ideal gas equation, PV = nRT P = (nRT)/(V) P = (1.5 × 0.0821 × 300)/(2) P = (36.945)/(2) = 18.4725\ atm The pressure of the gas is approximately 18.47\ atm . Answer: 18.47\ atm

MHT CET 202625 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 25 April
Shift: Morning Shift
Subject: Chemistry
Chapter: States of Matter

MHT CET 2026 · 25 April · Morning Shift · Chemistry · States of Matter

Calculate the pressure in atm of 48 g oxygen gas in a 2\ dm^3...

Calculate the pressure in atm of 48 g oxygen gas in a

2\ \text{dm}^3

container at

27\ ^{}^{\circ}\text{C}

.
[

R = 0.0821\ \text{atm dm}^3\ \textK^{-1}\ \text{mol}^{-1}

]

Options

A $18.47\ \text{atm}$
B $12.32\ \text{atm}$
C $6.16\ \text{atm}$
D $24.63\ \text{atm}$

Correct answer

A. $18.47\ \text{atm}$

Step-by-step solution

Given mass of oxygen gas,

w = 48\ \text{g}

Molar mass of oxygen gas (

O_2

M = 32\ \text{g mol}^{-1}

Number of moles,

n = \dfrac{w}{M} = \dfrac{48}{32} = 1.5\ \text{mol}

Volume of the container,

V = 2\ \text{dm}^3

Temperature,

T = 27\ ^{}^{\circ}\text{C} = 27 + 273 = 300\ \text{K}

Using the ideal gas equation,

PV = nRT

P = \dfrac{nRT}{V}

P = \dfrac{1.5 \times 0.0821 \times 300}{2}

P = \dfrac{36.945}{2} = 18.4725\ \text{atm}

The pressure of the gas is approximately

18.47\ \text{atm}

.

Answer:

18.47\ \text{atm}

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