Question

1) At standard temperature and pressure (0 ∘C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L. What volume would the same amount of gas occupy at the same pressure and 45 ∘C ?

Express your answer with the appropriate units.

2) One mole of an ideal gas is sealed in a 22.4-L container at a pressure of 1 atm and a temperature of 273 K. The temperature is then increased to 304 K , but the container does not expand. What will the new pressure be?

The most appropriate formula for solving this problem includes only which variables?

Enter the required variables, separated by commas (e.g., P,V,T).

3) If 3.0 g of N2 gas has a volume of 0.40 L and a pressure of 6.1 atm , what is its Kelvin temperature?

Express the temperature in kelvins to two significant digits.

Answer 1

1.At standard temperature and pressure (0 ∘C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L. What volume would the same amount of gas occupy at the same pressure and 45 ∘C ?

from gas law equation, P1V1/T1= P2V2/T2,where P1,V1 and T1 are initial conditions of pressure, temperature and volume , while P2, V2 and T2 are final conditions of temperature and pressure

given P1= 1 atm, V1= 22.4L, T1= 0+273=273K, P2= 1atm, V2= ?, T2= 45+273=318K

V2= V1T2/T1= 22.4*318/273 L=26.03 L

2.

given V1= 22.4 L, P1= 1 atm, T1=273K, T2= 304K, since the container does not expand, V2= 22.4L,

P2= P1V1*T2/T1V2= P1T2/T1= 1*304/273 =1.1135 atm

the variables are P,V and T connected by gas equation.

3. gas law equation can be written as PV=nRT

n= no of moles= mass of gas/molar mass of gas= 3/28 =0.107, V= 0.4L, P=6.1 atm, R=0.0821 L.atm/mole.K

T= PV/nR= 6.1*0.4/(0.107*0.0821)= 278 K