1. If 58.5 moles of an ideal gas is at 1.85 atm at 17.80°C, what is the volume o

Question

1. If 58.5 moles of an ideal gas is at 1.85 atm at 17.80°C, what is the volume of the gas?

2. If an ideal gas has a pressure of 7.33 atm, a temperature of 82.91°C, and has a volume of 62.97 L, how many moles of gas are in the sample?

3. At a certain temperature and pressure, one liter of CO2 gas weighs 1.55 g. What is the mass of one liter of O2 gas at the same temperature and pressure?

4. A 8.95-L container holds a mixture of two gases at 43°C. The partial pressures of gas A and gas B, respectively, are 0.355 atm and 0.861 atm. If 0.230 mol of a third gas is added with no charge in volume or temperature, what will the total pressure become?

Explanation / Answer

1) Apply the Ideal Gas Law :  

PV = nRT ,T=17.80 +273 K =290.8 K

V = ( n ) ( R ) ( T ) / (P )   

V = ( 58.5 mol ) ( 0.08206 atm - L / mol - K ) ( 290.8 K ) / ( 1.85 atm )

V = 1395.988308 /1.85

=> V = 754.6 L

2) Given P=7.33 atm , T =82.910C =82.91 +273 K =355.91 K, V =62.97 L

Now,apply the ideal gas law,

PV=nRT

=> n = PV /RT

=>n= 7.33 atm x 62.97 L /0.08206 L atm K-1 mol-1 x 355.91 K

=> n = 461.5701 /29.2059746 mol

=> n =15.8 mol

3) 1.55 * 32/44 = 49.6 /44 =1.127 g

Note P, V and T is the same so number of moles is the same

thus you scale by molecular weight of O2 = 32
over molecular weight of CO2 = 12+16+16 =44

4) Total pressure initially = sum of partial pressure of A and B = 0.355 atm + 0.861 atm = 1.216 atm

Now determne the initial moles of gas in the system from the total pressure

PV = nRT

P = 1.216 atm
V = 8.95 L
n = ? mol
R = gas constant = 0.08206 Latmmol-1K-1
T = temperature in Kelvin = 316 K

n = PV / RT
= 1.216 atm x 8.95 L / (0.08206 Latmmol-1K-1 x 316 K)
= 10.8832 /25.93096 mol

=0.4196 mol = moles of gas initially in system

New moles = 0.4196 mol + 0.230 mol
= 0.6496 mol


Now work out what the new pressure will be

P = ? atm
V = 8.95 L
n = 0.6496 mol
R = 0.08206 Latmmol-1K-1
T = 316 K

P = nRT / V
= 0.6496 mol x 0.08206 Latmmol-1K-1 x 316 K / 8.95 L

=16.84475162 /8.95
=> P= 1.88 atm

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