1. A container of gas at 4 atm pressure and 112 C is compressed at constant temp
ID: 485917 • Letter: 1
Question
1. A container of gas at 4 atm pressure and 112 C is compressed at constant temperature until the volume is halved. It is then further compressed at
a. What is the final pressure of the gas?
b. What is the final temperature of the gas?
2. On average, each person in the industrialized world is responsible for the emission of 10,000 kg of carbon dioxide (CO2) every year. This includes CO2 that you generate directly, by burning fossil fuels to operate your car or your furnace, as well as CO2 generated on your behalf by electric generating stations and manufacturing plants. CO2 is a greenhouse gas that contributes to global warming. If you were to store your yearly CO2 emissions in a cube at STP, how long would each edge of the cube be?
3. 0.66 mol of argon gas is admitted to an evacuated 60 cm3 container at 50 C. The gas then undergoes an isochoric heating to a temperature of 500 C. What is the final pressure of the gas?
Explanation / Answer
1.
A container of gas at 4 atm pressure and 112 C is compressed at constant temperature until the volume is halved. It is then further compressed at
1.What is the final pressure of the gas?
If T is constant then
p1 .V1= p2 . V2
p1 = 4 atm and V2 = V1 / 2
p2 = 8 atm
2.What is the final temperature of the gas?
if p constant then
V1 / T1 = V2 / T2
T1 = 385.15 K, V2 = V1 / 2
T2 = 192.57 K
final pressure = 8atm
final temperature = 195 K or -78.15 C
3]
Isochoric means constant volume.
Ideal gas law state 2:
P2*V = n*R*T2
P2 = n*R*T2/V
R is the molar universal gas constant, it is independent of flavor of gas.
From this we have
n=0.66 mol;
T2=773.15 K;
V=0.060 Liters;
R=8.314 kPa-Liter/mol-K;
P2 = n*R*T2/V
P2 = 0.66*8.314*773.15/0.060
P2 = 70707 kPa
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