The work done by a cretain material when temprature changes from `T_(0)` to `2T_(0)` m while pressure remains constant is `3betaT_(0)^(2)` where `beta` is a canstant. Draw curve between volume `(V)` and temeprature `(T)` of the material.

Calculate the work done in the following processes, when gas expands from V_0 to 2V_0 (a) P=kV , k is constant (b) T=T_0+alphaV , T_0 , alpha: constant n: number of moles of gas

Pressure and volume of a gas changes from (p_0V_0) to (p_0/4, 2V_0) in a process pV^2= constant. Find work done by the gas in the given process.

Pressure and volume of a gas changes from (p_0V_0) to (p_0/4, 2V_0) in a process pV^2= constant. Find work done by the gas in the given process.

A gas is undergoing an adiabatic process. At a certain stage, the volume and absolute temperature of the gas are V_(0), T_(0) and the magnitude of the slope of the V-T curve is m. molar specific heat of the gas at constant pressure is [Assume the volume of the gas is taken on the y-axis and absolute temperature of the gas taken on x-axis]

A gas is undergoing an adiabatic process. At a certain stage, the volume and absolute temperature of the gas are V_(0), T_(0) and the magnitude of the slope of the V-T curve is m. molar specific heat of the gas at constant pressure is [Assume the volume of the gas is taken on the y-axis and absolute temperature of the gas taken on x-axis]

Pressure P , Volume V and temperature T of a certain material are related by the P=(alphaT^(2))/(V) . Here alpha is constant. Work done by the material when temparature changes from T_(0) to 2T_(0) while pressure remains constant is :

Pressure P , Volume V and temperature T of a certain material are related by the P=(alphaT^(2))/(V) . Here alpha is constant. Work done by the material when temparature changes from T_(0) to 2T_(0) while pressure remains constant is :

The temperature of a fixed mass of an ideal gas is changed from T_(1)K to T_(2)K(T_(2) > T_(1)) . Calculate the work done if the change is brought about at (i) constant pressure (ii) constant volume. Hence find from the first law of thermodynamics, in which case heat absorbed will be greater.