JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 28)
In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by $$\Delta $$T. The
amount of heat absorbed by gas is (R is gas constant) :
$${1 \over 2}$$ KR$$\Delta $$T
$${1 \over 2}$$ R$$\Delta $$T
$${3 \over 2}$$ R$$\Delta $$T
$${2K \over 3}$$ $$\Delta $$T
Explanation
VT = K
$$ \Rightarrow $$ V$$\left( {{{PV} \over {nR}}} \right)$$ = k $$ \Rightarrow $$ PV2 = K
$$ \because $$ C = $${R \over {1 - x}} + $$ Cv (For polytropic process)
C = $${R \over {1 - 2}} + {{3R} \over 2}$$ = $${R \over 2}$$
$$ \therefore $$ $$\Delta $$Q = nC $$\Delta $$T
= $${R \over 2} \times \Delta $$T
$$ \Rightarrow $$ V$$\left( {{{PV} \over {nR}}} \right)$$ = k $$ \Rightarrow $$ PV2 = K
$$ \because $$ C = $${R \over {1 - x}} + $$ Cv (For polytropic process)
C = $${R \over {1 - 2}} + {{3R} \over 2}$$ = $${R \over 2}$$
$$ \therefore $$ $$\Delta $$Q = nC $$\Delta $$T
= $${R \over 2} \times \Delta $$T
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