Question

If two moles of an ideal gas at $546\ \text{K}$ occupy a volume of $44.8\ \text{L}$, what is the pressure of ideal gas at $546\ \text{K}$? 
($R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$) 

• A. $20\ \text{atm}$
• B. $0.2\ \text{atm}$
• C. $0.5\ \text{atm}$
• D. $2.0\ \text{atm}$

Hint:

Always convert all quantities into consistent units before applying the ideal gas equation.

Answer 1

The Correct Option is D

Step 1: Ideal gas equation.
The ideal gas equation is given by:
\[ PV = nRT \]
Step 2: Substitution of given values.
\[ P \times 44.8 = 2 \times 0.0821 \times 546 \]
Step 3: Calculation.
\[ P = \frac{2 \times 0.0821 \times 546}{44.8} \] \[ P = 2.0\ \text{atm} \]
Step 4: Conclusion.
Hence, the pressure of the gas is $2.0\ \text{atm$}.