(a) What is the gauge pressure in a 26.0°C car tire containing 4.50 mol of gas i

Question

(a) What is the gauge pressure in a 26.0°C car tire containing 4.50 mol of gas in a 30.5 L volume? (Give your answer to at least two decimal places. For this question, use R = 0.0821 L · atm/K/mol.) ?atm

(b) What will its gauge pressure be if you add 1.00 L of gas originally at atmospheric pressure and 26.0°C? Assume the temperature returns to 26.0°C and the volume remains constant. (Give your answer to at least two decimal places. For this question, use R = 0.0821 L · atm/K/mol.) ?atm

(note: I got the first part correct. So A) is 2.62 atm, I just do not know how to work part B)) Thank you! Please show your work.

Explanation / Answer

b)

no of moles of the gad that is being added = P*V/(R*T) (using the relation P*V = n*R*T ==> n = P*V/(R*T))

= 1*1/(0.0821*(26+273))

= 0.0407 moles

so no moles of final gas, n = 4.5 + 0.041

= 4.5407 moles

now Apply, P*V = n*R*T

P = n*R*T/V

= 4.5407*0.0821*(26+273)/30.5

= 3.654 atm

P_gauge = P - Po

= 3.654 atm - 1 atm

= 2.654 atm <<<<<<<<-----------Answer

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