Calculate the ratio of specific heats for nitrogen. Given that the specific heat of nitrogen at cinstant pressure `=0.236` cal `g^(-1) K^(-1)` and density at S.T.P. is `0.001234 g//c c`. Atmospheric pressure`=1.01xx10^(6) dyn e//cm^(2)` .

Calculate the molar specific heat at constant volume . Given : specific heat of hydrogen at constant pressure is 6.85 "cal" mol^(-1) K^(-1) and density of hydrogen = 0.0899 "g" cm^(-3) . One mole of gas = 2.016g, J=4.2xx10^(7) "erg" cal^(-1) and 1 atmosphere = 10^(6) "dyne" cm^(-2) .

Calculate the molar specific heat at constant volume . Given : specific heat of hydrogen at constant pressure is 6.85 "cal" mol^(-1) K^(-1) and density of hydrogen = 0.0899 "g" cm^(-3) . One mole of gas = 2.016g, J=4.2xx10^(7) "erg" cal^(-1) and 1 atmosphere = 10^(6) "dyne" cm^(-2) .

Calculate the molar specific heat at constant volume . Given : specific heat of hydrogen at constant pressure is 6.85 "cal" mol^(-1) K^(-1) and density of hydrogen = 0.0899 "g" cm^(-3) . One mole of gas = 2.016g, J=4.2xx10^(7) "erg" cal^(-1) and 1 atmosphere = 10^(6) "dyne" cm^(-2) .

Calculate the difference between two specific heats of 1 g of nitrogen. Given molecular weight of nitrogen = 28 and J = 4.2 times 10^7 erg cal^-1 .

Calculate the change in internal energy when 5g of air is heated from 0^(@) to 4^(@)C . The specific heat of air at constant volume is 0.172 cal g^(-1) .^(@)C^(-1) .

'Specific heat of iron is "0.106 cal.g"^(-1).^(@)C^(-1) '- what do you mean by this statement?

Calculate the two specific heats of nitrogen from the following data : gamma = c_(p)//c_(v) = 1.51 density of nitrogen at N.T.P = 1.234 g litre ^(-1) and J = 4.2 xx 10^(7) erg cal^(-1) .

Calculate the amount of heat absorbed by 500 g of ice to change into water of 20^@C . (specific heat capacity of water is 1 cal//g^@C )