A gas at normal temperature is suddenly compressed to one-fourth of its original volume. If = 1.5 , then the increase in the temperature of the gas in Kelvin is ( is the ratio of specific heats)

Step-by-step solution

Normal temperature is T₁ = 273 K. Initial volume V₁ = V . Final volume V₂ = V 4 . Since the gas is suddenly compressed, the process is adiabatic. The relation between temperature and volume for an adiabatic process is T V^ - 1 = constant . T₁ V₁^ - 1 = T₂ V₂^ - 1 T₂ = T₁ ( V₁ V₂ )^ - 1 Substituting the given values: T₂ = 273 ( V V/4 )^ 1.5 - 1 T₂ = 273 (4)^ 0.5 T₂ = 273 2 = 546 K. The increase in temperature is T = T₂ - T₁ . T = 546 - 273 = 273 K.