Here we have to compare the state of helium gas at STP and high temperature and low pressure.
At STP (standard condition of temperature and pressure) i.e. 273K temperature and 1 bar pressure. At STP helium gas will behave as a real gas.
At higher temperature and low pressure Helium will behave as an ideal gas.
The ideal gas conditions are developed on taking into account two factors: (i) the gas molecules are point of mass and having no volume. (ii) there is no existence of force of attraction between the molecules.
The deviation from ideal gas to the real gas depends upon the van der waals' interaction between the gas molecules. Now in low pressure and high temperature, we can ignore the volume and also the inter-molecular force of attraction. Thus the gas sample can behaved as ideal gas.
But at elevated pressure and low temperature i.e. STP the assumptions are not valid and it will behave as real gas.
<span>Answer:
1170 ml is the final volume of the sample of gas, if the pressure and moles in the container is kept constant.</span>
Answer:
Q= mc∆T(ice) + mLF(ice) + mc∆T(water) + mLV(water) + mc∆T(steam)
m=250 g = 0.25 kg = ¼ kg c(ice)= 2100 J/kg.K c(water)= 4200 J/kg.K LF(ice)= 333.7 kJ/kg LV(water)= 2256 kJ/kg c(steam)= 2080 J/kg.K
Explanation:
Q= ¼ × 2100 × (0°-(-30°)) + ¼ × 333700 + ¼ × 4200 × (100°-0°) + ¼ × 2256000 + ¼ × 2080 × (145°-100°)
Q= 15750 + 83425 + 105000 + 564000 + 23400
Q= 791575 J
