Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
Answer:
A. 6atm
Explanation:
Using pressure law equation:
P1/T1 = P2/T2
Where;
P1 = initial pressure (atm)
T1 = initial temperature (K)
P2 = final pressure (atm)
T2 = final temperature (K)
According to this question;
P1 = 3 atm
P2 = ?
T1 = 120K
T2 = 240K
Using P1/T1 = P2/T2
3/120 = P2/240
Cross multiply
240 × 3 = P2 × 120
720 = 120P2
P2 = 720/120
P2 = 6atm
