QUARTER FINAL STAGE 2023
Abuakwa State SHS
St Louis SHS
Adisadel College
QUESTION
Determine the volume occupied by 3.0 mol of an ideal gas at 4.0 atm and 27°C.
NOTE: 67.2 dm³ (3.0 mol × 22.4 dm³/mol) would be the volume at s.t.p., not at 4.0 atm and 27°C.
$V = (3.0)(0.0821)(300)/4.0$
$V = 18.47$ L
in its worked steps, but then states an unconnected, unsupported final answer of "67.2 dm³" without any derivation linking to that number.
ANSWER (CORRECTED): $\approx18.5$ dm³
SOLUTION:
$T=300$ K;
$V = \dfrac{nRT}{P}$
$V = \dfrac{(3.0)(0.0821)(300)}{4.0}$
$V \approx 18.5$
dm³ (L).
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PRACTICE QUESTIONS
1. QUESTION: Determine the volume occupied by 2.00 mol of an ideal gas at 27.0°C and 83.1 kPa.
ANSWER: 60.0 dm³ or $6.00\times10^{1}$ dm³
SOLUTION:
$V = \dfrac{nRT}{P}$
$T = 27.0 + 273$
$T = 300\text{ K}$
$V = \dfrac{2.00\times8.31\times300}{83.1}$
$83.1 = 10\times8.31$, so 8.31 cancels:
$V = \dfrac{2.00\times300}{10}$
$V = 2 \times 30$
$V = 60.0\text{ dm}^3$
2. QUESTION: Determine the volume occupied by 0.500 mol of an ideal gas at 127.0°C and 166.2 kPa.
ANSWER: 10.0 dm³ or $1.00\times10^{1}$ dm³
SOLUTION:
$V = \dfrac{nRT}{P}$
$T = 127.0 + 273$
$T = 400\text{ K}$
$V = \dfrac{0.500\times8.31\times400}{166.2}$
$166.2 = 20\times8.31$, so 8.31 cancels:
$V = \dfrac{0.500\times400}{20}$
$V = \dfrac{5\times10^{-1} \times 400}{20}$
$V = 5 \times 20 \times10^{-1}$
$V = 10.0\text{ dm}^3$
3. QUESTION: Determine the volume occupied by 3.00 mol of an ideal gas at 77.0°C and 249.3 kPa.
ANSWER: 35.0 dm³ or $3.50\times10^{1}$ dm³
SOLUTION:
$V = \dfrac{nRT}{P}$
$T = 77.0 + 273$
$T = 350\text{ K}$
$V = \dfrac{3.00\times8.31\times350}{249.3}$
$249.3 = 30\times8.31$, so 8.31 cancels:
$V = \dfrac{3.00\times350}{30}$
$V = \dfrac{3 \times 35}{3}$
$V = 35.0\text{ dm}^3$
4. QUESTION: Determine the volume occupied by 2.00 mol of an ideal gas at 127.0°C and 83.1 kPa.
ANSWER: 80.0 dm³ or $8.00\times10^{1}$ dm³
SOLUTION:
$V = \dfrac{nRT}{P}$
$T = 127.0 + 273$
$T = 400\text{ K}$
$V = \dfrac{2.00\times8.31\times400}{83.1}$
$83.1 = 10\times8.31$, so 8.31 cancels:
$V = \dfrac{2.00\times400}{10}$
$V = 2 \times 40$
$V = 80.0\text{ dm}^3$
5. QUESTION: Determine the volume occupied by 0.250 mol of an ideal gas at 227.0°C and 41.55 kPa.
ANSWER: 25.0 dm³ or $2.50\times10^{1}$ dm³
SOLUTION:
$V = \dfrac{nRT}{P}$
$T = 227.0 + 273$
$T = 500\text{ K}$
$V = \dfrac{0.250\times8.31\times500}{41.55}$
$41.55 = 5\times8.31$, so 8.31 cancels:
$V = \dfrac{0.250\times500}{5}$
$V = \dfrac{25\times10^{-2} \times 500}{5}$
$V = 5 \times 500 \times10^{-2}$
$V = 25.0\text{ dm}^3$
6. QUESTION: Determine the volume occupied by 0.600 mol of an ideal gas at 227.0°C and 33.24 kPa.
ANSWER: 75.0 dm³ or $7.50\times10^{1}$ dm³
SOLUTION:
$V = \dfrac{nRT}{P}$
$T = 227.0 + 273$
$T = 500\text{ K}$
$V = \dfrac{0.600\times8.31\times500}{33.24}$
$33.24 = 4\times8.31$, so 8.31 cancels:
$V = \dfrac{0.600\times500}{4}$
$V = \dfrac{6\times10^{-1} \times 500}{4}$
$V = 6 \times 125 \times10^{-1}$
$V = 75.0\text{ dm}^3$