Question

A sample of argon gas has a volume of 735 mL at a pressure of 1.20 atm and a temperature of 112 degrees C. What is the volume of gas, in milliliters, when the pressure and temerature of the gas sample are changed to each of the following? a) 658 mmHg and 281 K. b) 0.55 atm and 75 degrees C. c) 15.4 atm and -15 degrees C.

Answer 1

answer. plan, first find out the n(number of moles of argon gas)from the given data.

pressure P=1.20,volumeV=735/1000=0.735 L,temperature T=112+273=385 K.

as we know for ideal gas behaviour PV=nRT n=PV/RT-------(1) substitute all the values in eq (1).n=(1.2x0.735)/0.0821x385=0.88/31.61=0.0278 molesof argon gas.

a)P=658 mmHg=658/760=0.87atm,T=281 , V=nRT/P------(2) substitute the values in eq (2) V=(0.0278x0.0821x281)/0.87=0.64/0.87=0.74 L volume in ml=0.74x1000=740 ml.

b)P=0.55 atm T=75+273=348 K substitute the values in eq(2) V=(0.0278x0.0821x348)/0.55=0.79/0.55=1.44L volume in ml =1.44x1000=1440 ml.

b)P=15.4 atm, T=-15+273=258K. substitute the values in eq(2) V=(0.0278x0.0821x258)/ 15.4= 0.59=0.0383 L volume in ml =0.0383x1000= 38.3 ml