PRELIMINARY STAGE 2026
Contest 16
Tamale SHS: 68 points
Kalpohin SHS: 41 points
Wenchi Methodist SHS: 30 points
Contest 17
Kadjebi Asato SHS: 56 points
Ahantaman Girls’ SHS: 48 points
Ada SHTS, Sege: 35 points
Contest 18
Anglican SHS, Kumasi: 69 points
Namong SHTS: 28 points
Shama SHS: 25 points
ROUND 2 - SPEED RACE
QUESTION
A gas occupies 4.00 dm$^3$ at 300.0 K and 100.0 kPa.
What is the volume at 400.0 K and 200.0 kPa?
ANSWER: 2.67 dm$^3$
SOLUTION:
$V_2 = V_1 \times \dfrac{P_1}{P_2} \times \dfrac{T_2}{T_1}$
$V_2 = 4.00 \times \dfrac{100.0}{200.0} \times \dfrac{400.0}{300.0}$
$V_2 = 4.00 \times \dfrac{1}{2} \times \dfrac{4}{3}$
$V_2 = 2.00 \times \dfrac{4}{3}$
$V_2 = \dfrac{8}{3}$
$V_2 = 2.67\text{ dm}^3$
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PRACTICE QUESTIONS
1. QUESTION: A gas occupies 5.00 dm$^3$ at 200.0 K and 150.0 kPa. What is the volume at 400.0 K and 250.0 kPa?
ANSWER: 6.00 dm³
SOLUTION:
$V_2 = V_1\times\dfrac{P_1}{P_2}\times\dfrac{T_2}{T_1}$
$V_2 = 5.00\times\dfrac{150.0}{250.0}\times\dfrac{400.0}{200.0}$
$V_2 = 5.00\times\dfrac{3}{5}\times2$
$V_2 = 3.00\times2$
$V_2 = 6.00\text{ dm}^3$
2. QUESTION: A gas occupies 8.00 dm$^3$ at 320.0 K and 50.0 kPa. What is the volume at 240.0 K and 75.0 kPa?
ANSWER: 4.00 dm³
SOLUTION:
$V_2 = V_1\times\dfrac{P_1}{P_2}\times\dfrac{T_2}{T_1}$
$V_2 = 8.00\times\dfrac{50.0}{75.0}\times\dfrac{240.0}{320.0}$
$V_2 = 8.00\times\dfrac{2}{3}\times\dfrac{3}{4}$
$V_2 = 8.00\times\dfrac{1}{2}$
$V_2 = 4.00\text{ dm}^3$
3. QUESTION: A gas occupies 3.00 dm$^3$ at 300.0 K and 100.0 kPa. What is the volume at 400.0 K and 250.0 kPa?
ANSWER: 1.60 dm³
SOLUTION:
$V_2 = V_1\times\dfrac{P_1}{P_2}\times\dfrac{T_2}{T_1}$
$V_2 = 3.00\times\dfrac{100.0}{250.0}\times\dfrac{400.0}{300.0}$
$V_2 = 3.00\times\dfrac{2}{5}\times\dfrac{4}{3}$
$V_2 = 1.20\times\dfrac{4}{3}$
$V_2 = \dfrac{4 \times 12\times10^{-1}}{3}$
$V_2 = 4 \times 4 \times10^{-1}$
$V_2 = 1.60\text{ dm}^3$
4. QUESTION: A gas occupies 9.00 dm$^3$ at 300.0 K and 100.0 kPa. What is the volume at 200.0 K and 150.0 kPa?
ANSWER: 4.00 dm³
SOLUTION:
$V_2 = V_1\times\dfrac{P_1}{P_2}\times\dfrac{T_2}{T_1}$
$V_2 = 9.00\times\dfrac{100.0}{150.0}\times\dfrac{200.0}{300.0}$
$V_2 = 9.00\times\dfrac{2}{3}\times\dfrac{2}{3}$
$V_2 = 6.00\times\dfrac{2}{3}$
$V_2 = 2 \times 2$
$V_2 = 4.00\text{ dm}^3$
5. QUESTION: A gas occupies 4.00 dm$^3$ at 250.0 K and 200.0 kPa. What is the volume at 500.0 K and 100.0 kPa?
ANSWER: 16.0 dm³ or $1.60\times10^{1}$ dm³
SOLUTION:
$V_2 = V_1\times\dfrac{P_1}{P_2}\times\dfrac{T_2}{T_1}$
$V_2 = 4.00\times\dfrac{200.0}{100.0}\times\dfrac{500.0}{250.0}$
$V_2 = 4.00\times2\times2$
$V_2 = 8.00\times2$
$V_2 = 16.0\text{ dm}^3$
6. QUESTION: A gas occupies 12.0 dm$^3$ at 400.0 K and 90.0 kPa. What is the volume at 300.0 K and 60.0 kPa?
ANSWER: 13.5 dm³ or $1.35\times10^{1}$ dm³
SOLUTION:
$V_2 = V_1\times\dfrac{P_1}{P_2}\times\dfrac{T_2}{T_1}$
$V_2 = 12.0\times\dfrac{90.0}{60.0}\times\dfrac{300.0}{400.0}$
$V_2 = 12.0\times\dfrac{3}{2}\times\dfrac{3}{4}$
$V_2 = 18.0\times\dfrac{3}{4}$
$V_2 = \dfrac{3 \times 9}{2}$
$V_2 = 13.5\text{ dm}^3$