A gas thermometer is envisaged where by the change in V is used as a measure of temp change. At 1 atm what would the change in V as 42g of nitrogen gas goes from 15 to 20C? what difficulties can you envisage in using this system to measure room temp?

The volume of the gas will increase by a factor of ##1.073##.
The idea here is that you can use to calculate the change in volume caused by the given change in temperature, provided of course that the pressure stays constant.
The mass of nitrogen gas is not relevant here. This mass is equivalent to a number of moles of gas, but since the quantity of nitrogen gas remains unchanged, you don’t have to worry about exactly how many moles you have in there.
So, you know that when pressure and number of moles of gas are constant, volume and temperature have a direct relationship.
Simply put, when temperature increases, the volume of the gas Increases as well. SImilarlly, when temperature decreases, the volume of the gas decreases as well.
You can thus say that
##color(blue)(|bar(ul(color(white)(a/a)V_1/T_1 = V_2/T_2color(white)(a/a)|)))” “##, where
##V_1##, ##T_1## – the volume and absolute temperature of the gas at an initial state
##V_2##, ##T_2## – the volume and absolute temperature of the gas at a final stat
It’s very important to realize that you’re dealing with absolute temperature, which is temperature expressed in Kelvin, so make sure that you convert your two values before using the equation.
Your goal here is to solve for ##V_2##, the final volume of the gas
##V_1/T_1 = V_2/T_2 implies V_2 = T_2/T_1 * V_1##
Plug in your values to find
##V_2 = ( (273.15 + 20)color(red)(cancel(color(black)(“K”))))/((273.15 + 15)color(red)(cancel(color(black)(“K”)))) * V_1##
##V_2 = 1.073 * V_1##
You can thus say that the volume of the gas will increase by a factor of ##1.073## as a result of the temperature increasing from ##15^@”C”## to ##20^@”C”##.