it is found, on heating a gas its volume increases by 50% and pressure decreases to 60% of its original value. If the original temperature was - 15°C, find the temperature to which it was heated.
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The gas has constant mass as it is ideal. So we can use the equation for the general transformation of the ideal gas:
pV / T = const.
So pV / T = fpV(1+k)/T',
Where f = 0.6 and k = 0.5
It is found that T' = Tf (1+k) T = 273 - 15 = 258 K T'
= 258*0.6*1.5 = 232.2 K t’ = The temperature to which it was heated is -40.8 degrees Celsius.
Another way to write is: pV = nu*RT 0.6p*1.5V = nu*RT'
=> 0.9pV = nu*RT' So 0.9T = T'
=> T' = 232.2 K
This means that the gas is not heated.
pV / T = const.
So pV / T = fpV(1+k)/T',
Where f = 0.6 and k = 0.5
It is found that T' = Tf (1+k) T = 273 - 15 = 258 K T'
= 258*0.6*1.5 = 232.2 K t’ = The temperature to which it was heated is -40.8 degrees Celsius.
Another way to write is: pV = nu*RT 0.6p*1.5V = nu*RT'
=> 0.9pV = nu*RT' So 0.9T = T'
=> T' = 232.2 K
This means that the gas is not heated.
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