A sample of helium gas at a pressure of 853 mm Hg and a temperature of 47 °C, occupies a volume of 5.40 liters. If the gas is heated at constant pressure to a temperature of 73 °C, the volume of the gas sample will be __________ L.
Pressure of the gas P is 853 mm Hg = 853/760 atm = 1.122 as 1 atm = 760 mm Hg
Initial temperature of the gas Ti = 47 C = 273+47 K = 320 K
Initial volume Vi is 5.4 L
Pressure remains the same.
so, by ideal gas equation we have PVi = nRTi where n is the no. of moles of gas and R is the universal gas constant. So, Vi/Ti = nR/P --- Eqn1
If we increase the temperature of the gas to 73 C we have Tf = 273+73 K = 346 K and keep the pressure constant and assume that the final volume becomes Vf
then b ideal gas equation we have PVf = nRTf
so, Vf/Tf = nR/P ---- Eqn2
From Eqn1 & Eqn2 we have Vf/Tf = Vi/Ti , applying the values we get
Vf/346K = 5.4 L/320K so, Vf = 346*5.4/320 L = 5.838 L
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