From the gas laws, we can obtain that;
The final volume in the first question is 219 mL.
Gas laws stipulate the relationship between volume, pressure and temperature of gases based on the ideal gas approach.
1) V1 = 150 mL
P1 = 175 kPa
V2 = ?
P2 = 120 kPa
From Boyle's law;
P1V1 =P2V2
V2 = P1V1/P2
V2 = 175 kPa × 150 mL/120 kPa
V2 = 219 mL
2) V1 = 650 mL
T1 = 30°C or 303 K
V2 = ?
T2 = 0.0°C or 273 K
V1/T1 = V2/T2
V1T2 = V2T1
V2 = V1T2/T1
V2 = 650 mL × 273 K/303 K
V2 = 586 mL
3) P1 = 50.0 psi
T1 = 25.0°C or 298 K
P2 = ?
T2 = 100°C OR 373 K
P1/T1 = P2/T2
P1T2 = P2T1
P2 = P1T2/T1
P2 = 50.0 psi × 373 K/298 K
P2 = 63 Psi
4) V1 = 400.0 mL
V2 = ?
P1 = 600.0 mm Hg
P2 = SP = 760 mmHg
P1V1 = P2V2
V2 = P1V1/P2
V2 = 600.0 mm Hg × 400.0 mL/760 mmHg
V2 = 316 mL
5) For the lower range;
P2 = P1T2/T1
P2 = 30.0 psi × 253 K/300 K
P2 = 25.3 psi
For the higher range;
P2 = P1T2/T1
P2 = 30.0 psi × 323 K/300 K
P2 = 32.3 psi
The pressure lies between 25.3 psi and 32.3 psi
6) P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
V2 = P1V1T2/P2T1
V2 = 480 mL × 95.5 kPa × 273 K / 101.325 kPa × 310 K
V2 = 398 mL
7) PV = nRT
n = PV/RT
P = 105 kPa or 0.049 atm
V = 1.75-L
R = 0.082 atm L K-1mol1-1
T = 23°C or 296 K
n = 0.049 atm × 1.75-L/0.082 atm L K-1mol1-1 × 296 K
n = 0.0035 moles
8) PV = nRT
V = nRT/P
V = 1.5-mole × 0.082 atm L K-1mol1-1 × 308 K/ 1.12 atm
V = 33.8 L
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