Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa
Explanation:
Pressure in the submarine = 108.9 kPa
Volume, V = 2.4 * 10^5 L
Pressure, P = 116k Pa
Temperature, T = 312 K
Ideal gas law: PV = nRT or n = PV / RT
So, moles of gas, n =116 KPa * 2.4 * 10 ^5L / 8.314 LK Pa K^-1 *312 K
= 1.073 *10^4 mol
when temperature is changed to 293K,
PV = nRT or P = nRT / V
=1.073 *10^4 mol *8.314 LK Pa mol^-1 K^-1 *293 K / 2.4*10^5L
=108.9 K Pa
Pressure in the submarine when the temperature is changed to 293 K is 108.9 K Pa
I would say b and d but I wouldn’t really know :/
Initial Conditions:
Volume= v1= 417 cm³
Temperature= T1 = 278 K
Final Conditions:
Temperature= T2 = 231K
Volume = v2 =?
Use the general gas equation;
P1*v1/T1 = P2*v2/T2
As, the temperature is constant;
So,
v1/T1 = v2/T2
417/278 = v2/231
v2= 346.5 cm³
There are 2 moles in 8 grams