An ideal gas is in equilibrium at initial state with temperature T=137oC, pressure P = 0.75Pa and volume V = 0.75 m3. If there is a change in state in which the gas undergoes an isothermal process to a final state of equilibrium during which the volume is doubled. Calculate the temperature and pressure of the gas at this final state.

Question

anonymous · Answer

WKT  P*V=n*R*T

Since it is an isotheremal process  therefore, n, R , T are constant
Therefore temperature remains 137oC

Use (P_i)*(V_i) =(P_f)*(V_f)

(0.75)(0.75)=(P_f)*(1.50)

P_f = 0.35 Pa

anonymous · Answer

P=pressure
V=volume
n=no of moles
R=gas constant
T=temperature

anonymous · Answer

@shivam_bhalla  what about the final temperature

anonymous · Answer

Sorry P_f = 0.365 Pa

anonymous · Answer

Since it is an isothermal process,  therefore temperature remains constant