The correct option is A A→p,r,s B→q,t C→p,r,s D→q,r,t
If P=2V2, then using ideal gas equation, PV=nRT
⇒(2V2)(V)=nRT
or, nRT=2V3
or, T∝V3
Hence if volume ↑, temperature of gas will also increase.
Increase in temperature means dT is +ve.
or, dU∝dT
∴dU is also +ve.
Using first law of thermodynamics;
dQ=dU+dW
⇒dQ is also +ve
(∵dW is +ve for expansion of gas (V↑))
Hence heat is supplied during expansion of gas.
∴(A)→p,r,s
(B) If PV2=constant, using ideal gas equation PV=nRT
⇒(nRT)V=constant
VT=constant
V∝1T
Hence if volume of gas increases, its temperature will decrease.
For polytropic process,
C=CV+R1−x
C=3R2−R=R2
Heat supplied is dQ=nCdT
If temperature decreases dQ is negative
Now, for rise in temperature (dT→+ve) the volume decrease (V↓) or workdone by gas is -ve. However the heat supplied to gas in this case.
dQ∝dT
dT→+ve means dQ is also +ve.
⇒(B)→q,t
(C) Given, C=CV+2R
For polytropic process,
C=CV+R1−x
1−x=0.5
x=0.5
P√V=Constant
Using PV=nRT
T√V=constant
Hence with increase in temperature, V↑
Also if V↑ then dW is +ve and dU is +ve (T↑)
From relation (i), dQ=dU+dW
dQis +ve, if dU & dW are positive (when temperature of gas increases)
Hence heat is supplied (dQ→+ve) when gas is expanding [(V↑) or dW is +ve]
⇒(C)→p,r,s
(D) Given, CV−2R=C
C=CV+R1−x
1−x=−0.5
x=1.5
PV1.5=Constant
T√V=Constant
Hence with increase in temperature, the volume of gas decreases.
⇒T↑(dT is +ve) then V↓ (compression of gas), dW is -ve.
Also, dQ=n(CV−2R)dT
For monoatomic gas CV=32R i.e, C<0
Thus dQ is -ve when temperature increases, or we can say heat is supplied to gas (dQ→+ve) when its temperature decrease (T↑) or volume increases.
∴(D)→q,r,t
Why this question?Tip: It tests your analytical skills in relatingthe workdone by gas, change in temperature &volume by correlating the ideal gas equation(PV =nRT) with specific heat capacity of gasand first law of thermodynamics (dQ = dU + dW)