In a diesel engine cylinder compressed air from approximately standard pressure and temperature to about one – sixteenth the original volume and a pressure of about 50 atm. The temperature of the compressed air is
A) 225 K
B) 873 K
C) 970 K
D) 1043 K

Hint: We know that the state of air is gaseous, so to calculate the temperature when the air is compressed, we can use combined gas law (since pressure, temperature and volume, all are involved) and make a comparison before and after compression of air to obtain the required value.

Law to be used:
\[\dfrac{{PV}}{T} = K\] where K is any constant.
P, V, Tyre pressure, volume and temperature respectively.
As the temperature and pressure are standard, their value are:
Temperature = 273.15 K
Pressure = 1 atm

Complete step by step answer:
As the air is a gas, we can find the required value of temperature using gas law.
All the quantities – pressure, temperature and volume are involved, we can use combined gas law which provides the relationship between all these quantities.
The general equation of the law is:
\[\dfrac{{PV}}{T} = K\] where K is any constant.
For two different states of a system it can be given as:
\[\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}\] _________ (1)
According to the question:
Volume becomes one-sixteenth of the original 🡪 ${V_2} = \dfrac{1}{{16}}{V_1}$
Pressure becomes 50 times the original 🡪 ${P_2} = 50{P_1}$
The temperature initially was standard i.e 273.15 K 🡪 ${T_1} = 273.15K$
The temperature of the compressed air to be calculated is ${T_2}$
Substituting these values in (1), we get:
\[\dfrac{{{P_1}{V_1}}}{{273.15}} = \dfrac{{50{P_1} \times \dfrac{1}{{16}}{V_1}}}{{{T_2}}}\]
Calculating for ${T_2}$:
$
  {T_2} = \dfrac{{50}}{{16}} \times 273.15 \\
\implies {T_2} = 853.5K \approx 853K \\
 $
Therefore, the temperature of the compressed air is 853 K and the correct option is B).

So, the correct answer is “Option B”.

Note:
Standard temperatures in different units have different values like 0°C and 32°F but the options here for the second temperature were in kelvin, so we considered its Kelvin’s value.
The gas law used here is the combination of gas laws: Charle’s, Boyle, Gay Lussac and Avogadro’s laws.
Standard condition and normal conditions of temperature and pressure are called STP and NTP respectively.