1.)A sample of argon gas at a pressure of 0.633 atm and a temperature of 261 ° C
ID: 1039739 • Letter: 1
Question
1.)A sample of argon gas at a pressure of 0.633 atm and a temperature of 261 °C, occupies a volume of 694 mL. If the gas is heated at constant pressure until its volume is 796 mL, the temperature of the gas sample will be___________°C.
2.)0.962 mol sample of carbon dioxide gas at a temperature of 20.0 °C is found to occupy a volume of 21.5 liters. The pressure of this gas sample ismm ____________ Hg.
3.)The stopcock connecting a 3.19 L bulb containing argon gas at a pressure of 2.96 atm, and a 4.56 L bulb containing carbon dioxide gas at a pressure of 1.69 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is ________________atm.
Explanation / Answer
Answer:
(1) Given Pressure P=0.633 atm,
Initial volume V1=694 mL and
temperature T1=261°C=261+273 K=534 K
And at constant pressure, it's volume rises V2=796 mL at T2=?
From ideal gas equation, PV=nRT
V/T=constant
V1/V2=T1/T2 => T2=(V2 xT1)/V1
T2=(796 mL x 534 K)/694 mL=612.48 K.
T2=339.5°C. (Since 0°C=273 K)
(2) Given moles of CO2 n=0.962 mol, temperature T=20°C=20+273 K =293 K, volume V=21.5 L,
gas constant R= 62.36367 L mmHg mol^-1 K^-1
From ideal gas equation PV=nRT
P=nRT/V
P=(0.962 mol)x(62.36366 mmHg mol^-1 K^-1)x(293 K)/(21.5L)
P=817.59 mmHg.
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