Pressure of `1` mole ideal is given by `P-P_(0)[1-(1)/(2)(V_(0)/(V))^(2)]` ,brgt If volume of gas change from `V` to `2 V`. Find change in temperature.
Pressure of 1 mole ideal is given by P=P_(0)[1-(1)/(2)(V_(0)/(V))^(2)] ,brgt If volume of gas change from V_(0) to 2 V_(0) . Find change in temperature.
Pressure of an ideal gas is given by P=P_(0)=[1-(1)/(2)((V_(0))/(V))^(2)]
One mole of an ideal gas undergoes a process P = P_(0) [1 + ((2 V_(0))/(V))^(2)]^(-1) , where P_(0) V_(0) are constants. Change in temperature of the gas when volume is changed from V = V_(0) to V = 2 V_(0) is:
The pressure (P) and absolute temperature (T) of an ideal gas are related to each other as Pprop(1)/(T^(2)) . If temperature of gas is 300 K and its volume changes from V to 8 V, then find the final temperature of gas.
One mole of an ideal gas undergoes a process p=(p_(0))/(1+((V_(0))/(V))^(2)) . Here, p_(0) and V_(0) are constants. Change in temperature of the gas when volume is changed from V=V_(0) to V=2V_(0) is
One mole of an ideal gas passes through a process where pressure and volume obey the relation P=P_0 [1-1/2 (V_0/V)^2] Here P_0 and V_0 are constants. Calculate the change in the temperature of the gas if its volume changes from V_0 to 2V_0 .
For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation (dp)/(dv) =-ap If p = p_(0) at v = 0 is the given boundary condition, then the maximum temperature one mole of gas can attain is : (Here R is the gas constant)
For a sample of n mole of a perfect gas at constant temperature T (K) when its volume is changed from V_1 to V_2 the entropy change is given by
One mole of ideal gas goes through process P = (2V^2)/(1+V^2) . Then change in temperature of gas when volume changes from V = 1m^2 to 2m^2 is :
