Chemistry gas laws

Question

1. Mainting constant pressure, the volume of a gas is decreased from 37.5 L to 15.3 L by heating it. If the original temperature was 60.0°C, what was the new temperature?
2. If a sample of gas occupies 5.9 L at 35°C and 91.0 kPa, what volume will the same gas occupy at 102°C and 110 kPa?
3. If 51.5 moles of helium gas are in a 25.8 L container under 3.5 atmospheres of pressure, what is the temperature in the container in Kelvins? (R=0.0821 atm L/ mol K)

Answer 1

1). If you heat the gas under constant pressure, the volume should INCREASE, not decrease as stated in the problem. Let's look at what happens using the information given in the problem. Here we are dealing with Charles Law. V1/T1 = V2/T2

V1 = 37.5 L

T1 = 60ºC +273 = 333K

V2 = 15.4 L

T2 = ?

37.5 L/333K = 15.4 L/T2

T2 = 137K which is a LOWER temperature than the original of 333K. So, it's hard to see how the gas was heated.

2). P1 = 91.0 kPa; V1 = 5.9 L; T1 = 35+273 = 308K

P2 = 110 kPa; V2 = ?; T2 = 102+273 = 375K

P1V1/T1 = P2V2/T2 ... combined gas law

(91.0)(5.9L/308 = (110)(V2)/375

V2 = 5.94 L = 5.9 L (to 2 sig. figs.)

3). Ideal gas law : PV = nRT

Solving for T: T= PV/nR

T = (3.5 atm)(25.8 L)/(51.5 moles)(0.0821 atm L/mol K)

T = 21.36 K = 21 K (to 2 sig. figs.)