At standard temperature and pressure, the volume of a tire is 3.5L. What is the new pressure if the temperature outside is 296k and its weight causes the volume of the gas is 2.0 L?

Answer:

The new pressure is: 1.896 atm

Explanation:

At standard temperature and pressure, we have:

$P_1 = 1atm$

$T_1 = 273.15k$

$V_1 = 3.5L$

Outside, we have:

$T_2 = 296k$

$V_2 = 2.0L$

Required

Determine the new pressure

Using combined gas law, we have:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

This gives:

$\frac{1 * 3.5}{273.15} =\frac{P_2*2.0}{296}$

Solve for $P_2$

$P_2 = \frac{296 * 1 * 3.5}{273.15*2.0}$

$P_2 = \frac{1036}{546.30}$

$P_2 \approx 1.896 atm$

7 0

3 years ago

How many moles of ammonia are produced when 5.0 moles of hydrogen react with excess nitrogen? Use this balanced equation for the

wel

3 mol H₂ → 2 mol NH₃

5 mol H₂ → x mol NH₃

x=2*5.0/3=3.3

n(NH₃)=3.3 mol

7 0

3 years ago