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 degree celcius, find the temperature to which it was heated in celcius

Asked by abeshchakraborty6 | 23rd Feb, 2020, 08:54: AM

Expert Answer:

Given:
 
T1 = −15 °C
 
     = −15 + 273 = 258 K
 
T2 = ?
 
V1 = V
 
V2 = V + 50 over 100 straight V
    equals space straight V plus bevelled 1 half straight V

equals bevelled 3 over 2 straight V
 
P1 = P
 
P2 = 60 over 100 straight P
 
P2 = 3 over 5 straight P
 
Using ideal gas Law,
 
fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight T subscript 1 end fraction equals fraction numerator straight P subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction

fraction numerator straight P cross times straight V over denominator 258 end fraction space equals fraction numerator open parentheses begin display style bevelled 3 over 5 end style close parentheses straight P cross times open parentheses begin display style bevelled 3 over 2 end style close parentheses straight V over denominator straight T subscript 2 end fraction

straight T subscript 2 space equals space fraction numerator 258 cross times 3 cross times straight P cross times 3 cross times straight V over denominator 5 cross times 2 cross times straight P cross times straight V end fraction

space straight T subscript 2 equals space 232.2 space straight K
 
The temperature is 232.2 K

Answered by Varsha | 24th Feb, 2020, 12:25: PM