An ideal gas at temperature T_1 is compressed to 32th of its original volume, then its temperature T_2 will be _____ (gamma=1.4)

An ideal gas at temperature T_1 is compressed to 32th of its original

1.	An ideal gas at temperature T_1 is compressed to 32th of its original volume, then its temperature T_2 will be _____ (gamma=1.4)
Answer» SOLUTION :For an ADIABATIC PROCESS, `T_1V_1^(gamma-1)=T_2V_2^(gamma-1)` `THEREFORE T_2=T_1(V_1/V_2)^(gamma-1)` `=T_1(V_1/(V_1//32))^(gamma-1)=T_1(32)^(1.4-1)` `=T_1xx(2^5)^(2//5)` `=T_1xx4` `=4T_1`