Answer:
The pressure upstream and downstream of a shock wave are related as

where,
= Specific Heat ratio of air
M = Mach number upstream
We know that

Applying values we get

Similarly the temperature downstream is obtained by the relation
![(T_(1))/(T_(o))=([2\gamma M^(2)-(\gamma -1)][(\gamma -1)M^(2)+2])/((\gamma +1)^(2)M^(2))](https://img.qammunity.org/2020/formulas/engineering/college/ds01b3nqpzavtmc78wr36v53v0fwrl7tum.png)
Applying values we get
![(T_(1))/(423)=([2* 1.4* 1.8^(2)-(1.4-1)][(1.4-1)1.8^(2)+2])/((1.4+1)^(2)* 1.8^(2))\\\\\therefore (T_(1))/(423)=1.53\\\\\therefore T_(1)=647.85K=374.85^(o)C](https://img.qammunity.org/2020/formulas/engineering/college/t0wios9rvo8ctqafak263ewyvlo9yyc5kw.png)
The Mach number downstream is obtained by the relation
