Kelvi

Kelvi

0 pts

0days until NEET 2026Every question counts. ✊

🧠

Did you know?

The human brain has ~86 billion neurons, each connected to up to 7,000 others.

Kelvi

0 pts

A $10.0$ L flask contains $64 g$ of oxygen at $27^\circ$ C. (Assume O $_2$ gas is behaving ideally). The pressure inside the flask in bar is (Given $R = 0.0831$ L bar K $^{-1}$ mol $^{-1}$ )

medium

Solutions

2022

chemistry

$2.5$ bar

$498.6$ bar

$49.8$ bar

$4.9$ bar

Explanation

To solve this problem, we will use the ideal gas equation: $\\ PV = nRT \\$ Given: $\\$ • Volume $V = 10.0$ L $\\$ • Mass of oxygen $= 64$

Our AI powered practice platform can help you achieve your doctor dream.

Practice 2000+ previous year NEET questions
More practice, more score
AI generated hints
Use hints to get start solving
AI companion chat to clear doubts 24*7
Ask any question, get instant answers
AI generated solutions
Get detailed step by step solutions
Check related NCERT content
Read while solving
Track your progress
Improve every day

Sign up / Login

g

\\

• Temperature

T = 27^\circ

C

\\

• Universal gas constant

R = 0.0831

L bar K

^{-1}

mol

^{-1} \\

First, convert the temperature from Celsius to Kelvin:

\\ T = 27 + 273.15 = 300.15

K

\\

Next, calculate the number of moles of

O_2: \\

Molar mass of

O_2 = 32

g/mol

\newline n = \frac{mass}{molar \ mass} = \frac{64 g}{32 g/mol} = 2

mol

\\

Now, substitute the values into the ideal gas equation:

\\ P \cdot 10.0 = 2 \cdot 0.0831 \cdot 300.15 \\

Simplify the equation:

\\ P \cdot 10.0 = 49.86 \\ P = \frac{49.86}{10.0} \\ P = 4.986

bar

\\

Rounding to one decimal place, the pressure inside the flask is

4.9

bar.

\\

This corresponds to Option 4.