Part 3 (van der Waals) 7. A cylinder in a diesel engine compresses gas from 1 L
ID: 1043533 • Letter: P
Question
Part 3 (van der Waals) 7. A cylinder in a diesel engine compresses gas from 1 L to 50 mL. The air/diesel mixture entering the cylinder is at I atm. At maximum compression (50 mL), the pressure inside the cylinder is 67 atm. a) What is the temperature at the its most compressed volume? b) Take the van der Waals constant for air/diesel mixture to be a -1.38 and b 0.037. Assume that the ideal gas law is correct at the initial conditions of filling the cylinder, where the cylinder is 1 Land at l atm, and 25 °C. i) What pressure does the ideal gas law predict at full compression (50 mL) in the cylinder, given the temperature you calculated in a? n) What pressure does van der Wals equation predict at full compression (50 mL.) in the cylinder given the temperature calculated in a? ii) Which gas law gives a better prediction of gases in a diesel engine cylinder? Justify our answerExplanation / Answer
FROM IDEAL GAS LAW
PV = nRT
AT MOST COMPRESSED VOLUME= 50 mL
PRESSURE = 67 ATM
GAS CONSTANT R= 0.0821 litre atm MOL- K-
n= 1
T= PV/nR
= (67 ATM* 50 mL)/ (1 MOL* 0.0821 litre atm MOL- K-)
= (335 * 10-3 LATM)/( 0.0821litre atm K-)
4.08 K
2 GIVEN
a= 1.38 atm litre2/mol2
b= 0.031 ml /mol
FROM IDEAL GAS LAW
PV = nRT
AT MOST COMPRESSED VOLUME= 50 mL
PRESSURE = ? ATM
GAS CONSTANT R= 0.0821 litre atm MOL- K-
n= 1
temperature= 4.08k
p= nRT/V
= (1 MOL* 0.0821 litre atm MOL- K- * 4.08k )/50 mL
= 6.69 ATM
[ P+( a*(n/V)2)(V-nb)]= nRT
FOR n=1
[ P+( a/V2)(V-b)]= RT
P= [RT/(V-b)] - a/v2
= [(0.0821 litre atm K- * 4.08k )/(50 mL- 0.031 milli litre] - [1.38 atm mlitre2/(50 ml)2]
= [( 0.335 litre atm)/ 49.969 ml] - [ 1.38/2500 atm]
[6.70 atm - 0.0052 atm]
= 6.69 atm
since in question 2 ideal gas condition are used
1 atm pressure and 298.15k so pv= nRt is MOST VALID