Initially the pressure of 1 mole of an ideal gas is 10^5 Nm ⁻² and its volume is 16 litre. When it is adiabatically compressed, its final volume is 2 litre. Work done on the gas is [ . molar specific heat at constant volume .= 3 R 2 ]

Step-by-step solution

aligned & P₁=10^5 Nm ⁻², ~V ₁=16 lit, V ₂=2 lit, C _ v = 3 R 2 & C _ p = C _ v + R = ( 3 R 2 + R )= 5 R 2 & = C _ p C _ v = 5 R 2 3 R 3 R 2 = 5 3 aligned For adiabatic process, P₁ V₁^ =P₂ V₂^ P₂=P₁ ( V₁ V₂ )^ =10^5 ( 16 2 )^ 5 3 =32 10^5 Nm ⁻² Work done, W = P ₁ ~V ₁- P ₂ ~V ₂ -1 = 10^5 16 10⁻³-32 10^5 2 10⁻³ ( 5 3 -1 ) =-7.2 ~kJ (by the gas) Work done on the gas is 7.2 kJ