Question

1. An unknown volume of gas has a pressure of 365 mm Hg and a temperature of 52^oC. If the pressure is raised to 1.2 atm, the temperature decreased to 47^oC, and the final volume measured to be 4.8×10³ mL, what was the initial volume of the gas in liters?

a. A gas that has a volume of 28 liters, a temperature of 45°C, and an unknown pressure has its volume increased to 34 liters and its temperature decreased to 35°C. If you measure the pressure after the change to be 2.0 atm, what was the original pressure of the gas?

b. A glass jar containing 0.27 mol of Nitrogen, with a leaky lid is heated from room temperature (22 ºC) to 150 ºC. How many moles of Nitrogen remain in the jar? How many grams?

Answer 1

Q.No-01

From question

Initial pressure , P1 = 365 mmHg = 365/760 = 0.48 atm Final Pressure , P2 = 1.2 atm

Initial Temperature, T1 = 52 ^oC = 52 + 273 = 325 K, Final Temperature, T2 = 47 ^oC = 320 K

Suppose initial volume , V1 = ? Final Volume , V2 = 4.8 * 10³ ml = 4.8* 10³ / 10³ L = 4.8 L

We know the relationship from gas law

P1V1/T1 = P2V2/T2

So V1 = (P2V2T1) / P1T2

V1 = ( 1.2 atm * 4.8 L * 325 K ) / ( 0.48 atm * 320 K)

V1 = 1872 L / 153.6 = 12.18 L

Q.No- (a)

From question

Initial Volume , V1 = 28 L, Final Volume , V2 = 34 L

Initial Temperature, T1 = 45^oC = 45+ 273 = 318K , Final temperature , T2 = 35^oC = 35 + 273 = 308K

Suppose initial Pressure, P1 = ? Final pressure , P2 =2.0 atm

From gas law ,we know the relationship

P1V1/T1 = P2V2/T2

So P1 = (P2V2T1) / (V1T2)

P1 = ( 2 atm * 34L * 318K) / ( 28L * 308K)

P1 = ( 21624 atm) / 8624 = 2.507 atm